Chapter 3 Solutions 6nbd

>< > x=0 x<– 3 x=– 3 – 3 <x<0 f (x )′′ f(x)

f (x )′′ f(x)

Negative Concave down 0<x<+ 3 Negative Concave down

0 Point of inflection

Positive Concave up

x=+ 3 0 Point of inflection

x>+ 3 Positive Concave up

0 Point of inflection

 3 3  The points of inflection are  − 3, −  , and (0, 0) .  ,  3, 4  4   

MHR • Calculus and Vectors 12 Solutions

339

x−4 = x −2 (x − 4) 2 x −2 h′(x) = x (1) + (Ğ2)x −3 (x − 4)

e) h(x) =

= x −3 (x − 2x + 8) 8− x x3 h′(x) = x −3 (Ğ1) − 3x −4 (8 − x) =

= x −4 (−x + 3x − 24) =

2x − 24 x4

i) Take limits. x−4 x2 1 4 − 2 = lim x x x→∞ 1 =0

lim h(x) = lim

x→∞

x→∞

x−4 x→−∞ x 2 1 4 − 2 = lim x x x→−∞ 1 =0

Divide by x 2 /x 2 .

lim h(x) = lim

x→−∞

Divide by x 2 /x 2 .

ii) Examine the following. (−x) − 4 h(−x) = (−x)2 −x − 4 x2 The function is neither odd nor even. There is no resulting symmetry. =

iii) Find critical values. 8− x =0 x3 8− x = 0 x=8 Test using the second derivative test. 2(8) − 24 h′′(8) = (8)4 = −0.002


340

 1 Point  8,  is a maximum  16 

iv) Find possible points of inflection. 2 x − 24 =0 x4 2 x − 24 = 0 x = 12 There is 1 possible point of inflection.

h′′ ( x ) h(x)

x<0 Negative Concave Down

x =0 Undefined Vertical Asymptote

0 < x < 12 Negative Concave Down

x = 12 0 Point of Inflection

x > 12 Positive Concave Up

1  The point of inflection is 12,  . 18  

Chapter 3 Section 5

Question 19 Page 194

a) There is no x-intercept since the fraction can never equal zero. The y-intercept is 1 since 1 g(0) = 2 0 +1 =1 b) The maximum value for g(x) is 1. The function g(x) is a fraction whose numerator is fixed as 1. For a maximum value, the denominator must be as small as possible. Since x 2 is always greater than or equal to zero, the smallest the denominator can be is 0 + 1 = 1. Therefore, the maximum value for the fraction is 1. c) For large (positive or negative) values of x, g(x) becomes 1 over a very large number, which gets closer and closer to the value zero. The horizontal asymptote is the y-axis or y = 0 . d) No. The function has a positive y-intercept that is its maximum. As x gets large, the function decreases and approaches the x-axis from above. It is very unlikely that there would be any extrema in this situation. (However, it might be possible to imagine a damped sinusoidal curve that could satisfy the conditions of parts a) to c) of this question, but still have multiple local maxima and minima as the function approaches the x-axis.) Clearly, using the tools of calculus (e.g., derivatives) would this conjecture. e) Two. The function has one maximum where the function must be concave down. At the limit, the function must be concave up (or it will intersect the axis). There must be one point of inflection between the maximum and the end of the x-axis in either direction.


341

f)

Chapter 3 Section 5


a)

b)

x 2 + 2x + 3 x 2 + 2x + 1+ 2 = x +1 x +1 2 x + 2x + 1 2 = + x +1 x +1 2 (x + 1) 2 = + x +1 x +1 2 = x + 1+ x +1

Another method is to use a CAS.


342

c) The two functions are s(x) = x + 1 and t(x) =

2 . x +1

The function t has a vertical asymptote; therefore f will also have the same vertical asymptote. Function t approaches the value zero as x increases; therefore f will approach the curve s as x increases. That is, s will be a slant asymptote for f. Chapter 3 Section 5


C is the correct answer. y = (x + 2)5 (x 2 − 1)4 dy = (x + 2)4 (x 2 − 1)3 (13x 2 + 16x − 5) dx

dy 0. = 0 , x = −2 or ±1, or two other real roots from the quadratic equation 13 x 2 + 16 x − 5 = dx dy With the exception of −2, will have a sign change when ing through each of these roots. Hence, dx there are four local maximum or minimum points.

When

Chapter 3 Section 5


D is the correct answer. 1 1 1 x2 , , , and y = are discontinuous at x = 0 . y = = y 2 x x x x Hence, their derivatives are also discontinuous at x = 0 . However, x = 0 is not in the domains of these functions. dy 1 The function y = x has derivative , which is discontinuous at x = 0 . = dx 2 x In this case, x = 0 is in the domain of the function.

The functions y =


343

Chapter 3 Section 6

Optimization Problems

Chapter 3 Section 6

Question 1 Page 201

For a maximum, find the first derivative and let it equal zero to find the critical points. h(t) = −4.9t 2 + 19.6t + 2 h′(t) = −9.8t + 19.6 0 = −9.8t + 19.6 t=2

Since this function is a parabola opening down, this critical point is a maximum. h(2) = −4.9(2)2 + 19.6(2) + 2 = 21.6

The maximum height of the ball is 21.6 m. Chapter 3 Section 6

Question 2 Page 201

Let the integers be represented by x and 20 − x and P be their product (the quantity to be maximized). P(x) = x(20 − x) = 20x − x 2 P′(x) = 20 − 2x

Find the critical points. 20 − 2 x = 0 x = 10 Since this function is a parabola opening down, this critical point is a maximum. The two integers are 10 and (20 – 10) = 10. Chapter 3 Section 6

Question 3 Page 201

Find the critical points. N (t) = −0.05t 2 + 3t + 5 N ′(t) = −0.1t + 3 0 = −0.1t + 3 t = 30

Since this function is a parabola opening down, this critical point is a maximum. It takes 30 years of experience to reach maximum productivity.


344

Chapter 3 Section 6

Question 4 Page 201

a) Let the length of each small pen be x metres. This uses 4x of fencing. 1200 – 4x of fencing remains for the two long sides. Each long side has length: 1200 − 4 x = 600 − 2 x . 2 Let A be the area enclosed.

x

x

x

x

A(x) = x(600 − 2x) = 600x − 2x 2

Find the critical points. A′(x) = 600 − 4x 0 = 600 − 4x x = 150

Since this function is a parabola opening down, this critical point is a maximum. A(150) = 600(150) − 2(150)2 = 45 000

The maximum area that can be enclosed is 45 000 m2. b) This condition would limit the function to the interval: 180 < x < 210 . (The right hand limit occurs because the long sides of the big pen must also be at least 180 m long and that would leave only 840 m for the four shorter sides.) Consequently, there would be no critical point in the interval. A maximum value would have to occur at an endpoint. Since A(180) = 43 200 and A(210) = 37 800 , the maximum possible area would be 43 200 m2 in this case.


345

Chapter 3 Section 6

Question 5 Page 201

Let the width of the square be x metres. Let l be the length of the combined pen.

l

x

x

x

l 3 x + 2l = 60 60 − 3 x l= 2

Let A be the overall area.  60 − 3x  A(x) = x   2  = 30x −

3 2 x 2

Find the critical points. A′(x) = 30 − 3x 0 = 30 − 3x x = 10

Since this function is a parabola opening down, this critical point is a maximum. 3 A(10) = 30(10) − (10)2 2 = 150 60 − 3(10) 2 = 15

l=

The maximum area possible is 150 m2 and the dimensions that give this area are 10 m by 15 m.


346

Chapter 3 Section 6

Question 6 Page 201

a) Let the dimensions of the showroom be x and y.

glass x y xy = 500 500 y= x

Let the total cost be represented by C.  500   500  C(x) = 600  + 2x + 1200  x   x  900 000 x = 2400x + 900 000x −1

= 2400x +

Find the critical points. C ′(x) = 2400 − 900 000x −2 0 = 2400 − 900 000x −2 0 = 2400x 2 − 900 000 x 2 = 375 x = ± 375 x = ±5 15

Considering the context, 0 < x < 500 . Check the total cost at x = 5 15 and the endpoints.

( )

000 ( ) 900 (5 15 )

C 5 15 = 2400 5 15 + = 92 952 C(1) = 2400(1) +

900 000 (1)

= 902 400 C(500) = 2400(500) +

900 000 (500)

= 1 200 000


347

The minimum cost occurs when the dimensions are 5 15 and

500 20 15 . = 3 5 15

The showroom should be about 19.4 m by 25.8 m. b) Roofing will have a fixed cost since its area (500 m2) does not change as different showroom rectangles are considered. Chapter 3 Section 6

Question 7 Page 201

a) Let the perimeter be P. Then P = 2h + 2r + πr and so h =

1 (P – 2r – πr). 2

πr 2 + 2rh 2 πr 2  P − 2r − πr  = + 2r    2 2

A=

πr 2 + Pr − 2r 2 − πr 2 2 A′ = πr + P − 4r − 2πr A′ = P − 4r − πr =

P − 4r − πr = 0 r=

P 4+π

1 (P − 2r − πr) 2 1  P   P  = P − 2  − π   4 + π    4 + π 2

h=

=

P  2 + π 1 −  4+ π 2

=

P  4+ π 2 + π −   2  4+ π 4+ π

P 2    2  4+ π P = 4+π =

So the ratio is: P h 4+π = P r 4+π =1


348

b) Let A be the area and assume the cost for the straight edges is $x/m. πr 2 A= + 2rh 2 1  πr 2  h= A− 2r  2  =

A πr − 2r 4

C(x) = x(2h + 2r) + 3x(πr)  A πr  = 2x  −  + 2xr + 3xπr  2r 4   A πr  = x  − + 2r + 3πr  r 2   −A π  C ′(x) = x  2 − + 2 + 3π    r 2  −A 5π  + 2 0 = x 2 +  r  2 A 5π +2 = 2 r2  5π  + 2 r 2 A=   2 h=

1  5π   + 2 r 2     2r  2

r  5π  πr + 2 −   4 2 2 πr 5πr +r− = 4 4 = πr + r =

= r (π + 1)

So the ratio is: h r (π + 1) = r r = π +1


349

Chapter 3 Section 6

Question 8 Page 201

a) Let the radius, height, and surface area be represented by r, h, and S.A. respectively.

r

SA

h

S.A. is the variable to be maximized. Need to eliminate one of the other variables. The volume is 1. 1 = π r 2h 1 h= 2 πr The surface area is: S.A. = 2π r 2 + 2π rh  1  = 2π r 2 + 2π r  2   πr  2 r 2 = 2π r + 2r −1 = 2π r 2 +

For a minimum, find the critical points. S.A.′ = 4π r − 2r −2 0 = 4π r − 2r −2 0 = 4π r 3 − 2 1 2π  r B 0.54 r=

3


350

To this is a minimum, test values that are close. S.A.(0.54) = 5.54 S.A.(0.5) = 5.57 S.A.(0.6) = 5.60

h=

1 πr2 3

= =

4π 2

π 3

4

π

The surface area is a minimum when r =

3

1 and h = 2π

3

4

π

or, in decimals, when

r  0.54 and h  1.08 .

b)

h h = d 2r 1.08 = 2 0.54

( )

=

1 1

The ratio of height to diameter is 1:1. c) No. The diameter of a pop can is determined by the size of the hand that will hold it, irrespective of volume and cost of materials. Regular pop cans have a 2:1 ratio. Other types of drink cans may have different ratios depending on the volume of their contents. Some mini pop cans have a ratio close to 1:1. Chapter 3 Section 6

Question 9 Page 201

Let the radius, height, and total cost be represented by r, h, and C respectively. C is the variable to be minimized. Need to eliminate one of the other variables. The volume is 500 cm2. V = π r 2h 500 = π r 2 h h=

500 πr2


351

The cost function is:

(

C(r) = (0.4)π r 2 + (0.2) π r 2 + 2π rh

)

  500   = (0.4)π r 2 + (0.2)  π r 2 + 2π r  2    πr    200 r 2 = 0.6π r + 200r −1 = 0.6π r 2 +

For a minimum cost, find the critical points. C ′(r) = 1.2π r − 200r −2 0 = 1.2π r − 200r −2 0 = 1.2π r 3 − 200 500 3π r B 3.76  r=

3

To this is a minimum, test values that are close. C(3.76) = 79.84 C(3.7) = 79.86 C(3.8) = 79.85 5 3 36π 2 500 cm and h = cm or, in decimals, π 3π when r  3.76 cm and h = 11.27 cm .

The minimum occurs when r = 3

Chapter 3 Section 6


a) Let the radius, height, and volume be represented by r, h, and V respectively. V = π r 2h The surface area is 1 m2. S.A. = π r 2 + 2π rh 1 = π r 2 + 2π rh h=

1− π r 2 2π r

 1− π r 2  V = πr2    2π r  =

r − πr3 2


352

b) For maximum volume, find the critical points. 1 V ′(r) = 1− 3π r 2 2 1 0 = 1− 3π r 2 2 1 r2 = 3π r=±

(

)

(

)

1 3π

3π 3π  r B ±0.33 r=±

Radius must be positive so discard –33 m. The maximum volume drum will occur when the radius is approximately 0.33 m. c)

( x ∈ 0, 1 , Xscl = 0.1, y ∈ 0, 0.2  , Yscl = 0.01 ) d) If the domain of r is restricted to 0 < r < 0.2 , then there are no critical values to consider. The maximum volume can only occur at one of the endpoints. In fact, since this function is increasing on this interval (as is evident from the graph), the maximum volume will occur at the end of the interval, i.e., when r = 0.2 . (0.2) − π (0.2)3 2 = 0.1− 0.004π

V (0.2) =

The maximum volume in this case will be 0.087 m3. Chapter 3 Section 6


Let the radius, height, and volume of the paper cylinder be represented by r, h, and V respectively. V is the variable to be maximized. V = π r 2h Need this equation to only involve two variables. The perimeter is to be 100 cm. 100 = 2h + 2(2π r) r=

50 − h 2π


353

Therefore, the formula becomes: 2

 50 − h  V (h) = π  h  2π  =

1 3 100 2 2500 h − h + h 4π 4π 4π

For maximum volume, find the critical points. 3 2 200 2500 V ′(h) = h − h+ 4π 4π 4π 3 2 200 2500 0= h − h+ 4π 4π 4π 2 0 = 3h − 200h + 2500

(

)(

0 = 3h − 50 h − 50

h=

)

50 , h = 50 3

Check the volume at the critical values. 1 100 2500 V (50) = (50)3 − (50)2 + (50) 4π 4π 4π =0 3

2

 50  2500  50  1  50  100  50  − + V = 4π  3  4π  3   3  4π  3  = 1473.7

 50  100 − 2   50  3  = 100 If one dimension of the paper is , then the other dimension is 2 3 3 50 100 The maximum volume occurs when the paper is cm by cm . 3 3

Chapter 3 Section 6


Let n be the number of additional trees planted and P be the total crop. The number of trees is now 50 + n and the production per tree becomes 200 − 5n . P = (50 + n)(200 − 5n) To maximize the crop, find the critical points. P′(n) = (50 + n)(−5) + (1)(200 − 5n) Product rule. 0 = (50 + n)(−5) + (1)(200 − 5n) 0 = −250 − 5n + 200 − 5n n = −5

The optimal number of trees to be planted is 50 – 5 = 45. MHR • Calculus and Vectors 12 Solutions

354

Note: Increasing the number of trees above 50 actually reduces the total crop. P(45) = 10 125 P(50) = 10 000 P(55) = 9 625

Chapter 3 Section 6


a) R(x) = (30 + x)(5000 − 100x) = −100x 2 + 2000x + 150 000

b) −30 < x < 50 Answers may vary. For example: Assume that lowering the price increases the number of attendees. When x = −30 , the ticket price is $0, which is clearly impractical. If x = 50 , then the number of attendees becomes zero which is also impractical. c) Find the critical points. R′(x) = −200x + 2000 0 = −200x + 2000 x = 10

To that this is a maximum, check values of R(x).. R(10) = 160 000 R(9) = 159 900 R(11) = 159 900

A ticket price of $40 will maximize the revenue from the concert. d) The $40 price results in 4000 people attending the concert. If only 1200 people can attend, the price should be set at $68 (a $38 price increase leads to a reduction of 3800 people). Any further increases in price will lead to a lower total revenue. R(38) = 81, 600 R(39) = 75 900 R(40) = 70 000


355

Chapter 3 Section 6


Let (x, 9, –x2) be any point on the parabola. The resulting inscribed rectangle has width 2x and height 9 − x 2 .

y

(x, 9 − x ) 2

x Let A represent the area. A(x) = 2x(9 − x 2 ) = 18x − 2x 3

For a maximum, find the critical points A′(x) = 18 − 6x 2 0 = 18 − 6x 2 x= 3

Check that this is a maximum. A

( 3)= 12

3

A(2) = 20 A(1) = 16

The rectangle with the largest area has area 20.8 units2.


356

Chapter 3 Section 6


a) To minimize fuel cost per kilometre, examine the critical points. 25 v + C(v) = 100 v 1 v + 25v −1 = 100 1 C ′(v) = − 25v −2 100 1 0= − 25v −2 100 0 = v 2 − 100(25) v = ± 2500 v = ±50

The negative value makes no sense for this problem. Check that v = 50 is a minimum. C(50) = 1.00 C(40) = 1.025 C(60) = 1.017

The speed that results in the lowest cost per kilometre is 50 km/h. b) The total cost function, T, is made up from fuel cost and driver cost.  v  1000  25  T (v) =  +  (1000) + 40   100 v   v  = 10v + 25 000v −1 + 40 000v −1 = 10v + 65 000v −1 T ′(v) = 10 − 65 000v −2 0 = 10 − 65 000v −2 0 = 10v 2 − 65 000 v = ± 6500  v B ±80.6

The negative value makes no sense for this problem. Check that v = 80.6 is a minimum. C(80) = 1612.50 C(80.6) = 1612.45 C(81) = 1612.47

The speed that results in the lowest total cost for the 1000-km trip is about 80.6 km/h


357

Chapter 3 Section 6


Let the number be x and the sum to be minimized be S. 1 S(x) = x 2 + x 2 = x + x −1 Examine the critical values. S ′(x) = 2x − x −2 0 = 2x − x −2 0 = 2x 3 − 1 x = 3 0.5  x B 0.794

Check that x = 0.794 is a minimum. S(0.794) = 1.8899 S(0.85) = 1.899 S(0.75) = 1.8958

The number is 3 0.5  0.794 . Chapter 3 Section 6


Brenda’s costs for the trip include the running costs for every 100 km (15 since she is travelling 1500 km) of the vehicle and the costs of paying herself.

(

)

30(1500) v 45Ê000 = 13.5 + 0.024v 2 + v

Ctot (v) = 15 0.9 + 0.0016v 2 +

Find the derivative of the total cost and set it to zero to find the speed that will minimize cost.

45Ê000 v2 45Ê000 0 = 0.048v − v2 0 = 0.048v 3 − 45Ê000

Ctot ′ (v) = 0.048v −

0.048v 3 = 45Ê000 

v 3 = 937Ê500 v =&97.9

The speed 97.9 km/h (approximately) will minimize Brenda’s costs.


358

Chapter 3 Section 6


Choose coordinate axes so that the plexiglass is a semicircle with equation = y Each point on the semicircle determines a possible rectangle.

1 − x 2 as shown.

The resulting inscribed rectangle has width 2x and height 1 − x 2 .

y

(x,

1 − x2

)

x Let A represent the area. A(x) = 2x 1− x 2 1

= 2x(1− x 2 ) 2

For a maximum, find the critical points: 1 1 −  1 A′(x) = 2x   (1− x 2 ) 2 (−2x) + 2(1− x 2 ) 2  2 −

1

= (1− x 2 ) 2 (−2x 2 + 2 − 2x 2 ) = (2 − 4x 2 )(1− x 2 ) =

−

1 2

2 − 4x 2 1 2 2

(1− x ) 0=

2 − 4x 2 1

(1− x 2 ) 2 x2 =

1 2

x = ± 0.5  x B ±0.71

The maximum point is confirmed by the graph of A(x):.

QuickTime™ and a TIFF (LZW) decompressor are needed to see this picture.

The largest rectangle has dimensions 1.42 m by 0.71 m. MHR • Calculus and Vectors 12 Solutions

359

Chapter 3 Section 6


a) If x m are used for the quarter circle, then there are (20 – x) m for the two sides of the square.  20 − x  Each side will be  m.  2  b) 4x = 2π r 4x r= 2π 2x r=

π

c) A(x) = area of square + area of quarter circle 2

 20 − x  1  2x  = + π   4 π   2 

2

d) Simplifying, 400 − 40x + x 2 x 2 A(x) = + 4 π  1 1 2 = 100 − 10x +  +  x 4 π  4+π  2 = x − 10x + 100  4π 

Chapter 3 Section 6


a) Draw a diagram.

b) V (x) = x(40 − 2x)(60 − 2x) = 4x 3 − 200x 2 + 2400x


360

c) 0 < x < 20 If x < 0 , no tin would be cut from the corners and the box would have no volume. If x > 20, all of the tin is cut away from the width and no box can be formed. d) Find the critical values. V ′(x) = 12x 2 − 400x + 2400 0 = 12x 2 − 400x + 2400 0 = 3x 2 − 100x + 600 x=

100 ± 10000 − 7200 6

x=

100 ± 20 7 6

50 ± 10 7 3 x 25.5, x 7.85 x=

Check values for the one critical value in the domain for x. V (7.85) = 8450.4 V (7) = 8372 V (9) = 8316

Therefore: 50 − 10 7 3 B 7.85  x=

20 + 20 7 3 B 24.3

80 + 20 7 3 B 44.3

40 − 2x =

60 − 2x =





The box dimensions that will maximize the volume are 7.85 cm by 24.3 cm by 44.3 cm. Chapter 3 Section 6


a) Find the volume and surface area. V = π r 2 (2r) S.A. = 2π r 2 + 2π r(2r) = 2π r 3

= 6π r 2

The profit function is P(r) = 7(6π r 2 ) − 10(2π r 3 ) . = 42π r 2 − 20π r 3

To maximize the profit, find the critical values. P′(r) = 84π r − 60π r 2 0 = 84π r − 60π r 2 0 = 12π r(7 − 5r) r = 0, r = 1.4


361

Check values for this critical point. P(1.4) = 86.21 P(1.5) = 84.82 P(1.3) = 84.95

The container radius that maximizes the profit is 1.4 m. b) The profit per container is $86.21 c)

Chapter 3 Section 6


a) A cube is the most efficient shape. Non-cubical shapes require more material to enclose the same volume. b) No. The volume involved does not affect the efficient shape considerations. c) The shape would be expected to change. Let the volume be 1 m3. Also let the base be x by x and the height be h, V = (x)(x)(h) 1 = (x)(x)(h) h=

1 x2

The surface area is  1 S.A. = x 2 + 4x  2  x  = x 2 + 4x −1

For a minimum surface area, find the critical values. S.A.′ = 2x − 4x −2 0 = 2x − 4x −2 0 = 2x 3 − 4 x3 = 2 x= 32  x B1.26


362

The corresponding height is: 1 h= 2 3 2

( )

B 0.63

If the box has no top, the most efficient shape is a square base with a height that is one-half the base dimensions. Chapter 3 Section 6


a) i) Let the dimensions be x and y as shown.

y x There is 1000 m of fencing. 2x + 2 y = 1000 y = 500 − x A(x) = x(500 − x) = 500x − x 2

For a maximum: A′(x) = 500 − 2x 0 = 500 − 2x x = 250

The dimensions of the pen with maximum area are 250 m by 250 m. ii)

y x There is 1000 m of fencing, 2x + 3y = 1000 1000 − 2x y= 3


363

 1000 − 2x  A(x) = x   3  =

2 1000 x − x2 3 3

For a maximum: 1000 4 A′(x) = − x 3 3 1000 4 − x 0= 3 3 x = 250 The dimensions of the pen with maximum area are 250 m by 166.7 m. iii)

y x There is 1000 m of fencing. x + 2 y = 1000 1000 − x y= 2  1000 − x  A(x) = x   2  = 500x −

1 2 x 2

For a maximum, A′(x) = 500 − x 0 = 500 − x x = 500

The dimensions of the pen with maximum area are 500 m by 250 m. b) The dimension with two sections of fence is always 250 m. The other dimension is 500 divided by the number of fence sections in that dimension.


364

c) The dimensions, if there were three dividers and four equal parts, would be 250 m by 100 m. This result could be found by modifying the calculus technique used in part a). It could also be found by using the pattern identified in part b).

y x Chapter 3 Section 6


a) The cube has the least surface area. This can be verified using calculus, assuming the base is a square. Let the volume be 1 m3. Also let the base be x by x and the height be h, V = (x)(x)(h) 1 = (x)(x)(h) h=

1 x2

The surface area is  1 S.A. = 2x 2 + 4x  2  x  = 2x 2 + 4x −1

For a minimum surface area, find the critical values. S.A.′ = 4x − 4x −2 0 = 4x − 4x −2 0 = 4x 3 − 4 x3 = 1 x= 31 x =1

The corresponding height is 1. The dimensions are 1 m by 1 m by 1 m, a cube. b) It appears that the best shape is one that has all dimensions equal. In three-space, that shape is a sphere. Compare a cube and a sphere. Let the volume be 1 m3. For the cube, the dimensions are 1 m by 1 m by 1 m and the total surface area is 6 m2.


365

For the sphere: 4 1 = πr3 3 3 4π r B 0.62  r=

3

S.A. = 4π (0.62)2 

B 4.84

Clearly, the sphere has a smaller surface area than a cube of the same volume. c) Answers may vary. For example: : • Packaging may not be a significant cost related to the product. • Packaging is advertising. A larger surface pointed at the consumer provides the company with a l space in which to promote their product. • Spheres are very hard to stack or store on rectangular shelves. • Packages need to be handled. The size of the human hand may determine the thickness of the box. • Aesthetically, a rectangle in the shape of the golden ratio is more pleasing than a square. Chapter 3 Section 6


The correct answer is E. Let r be the radius of the circle, where 0 ≤ r ≤ The side length of the square is

L − 2π r . 4

L . 2π

The total area is:  L − 2π r  = A π r2 +   4  

2

dA π (π + 4)r π L = − dr 2 4

dA L < 0 when 0 ≤ r < dr 2(π + 4) dA L L . > 0 when
Therefore, A has a minimum when r = Hence 2π r =

L . 2(π + 4)

πL is used for the circle. π +4


366

Chapter 3 Section 6


The correct answer is E. f (x) = f ′(x) =

xn x −1 x n−1 (n − 1)x − n  (x − 1)2

If n is even, then f will have a local maximum at x = 0 and a local minimum at x = If n ≠ 1 is odd, then f will have only a local minimum at x =

n . n −1

n . n −1

If n = 1, then f will have no local maximum or minimum.


367

Chapter 3 Review Chapter 3 Review a)

Question 1 Page 204

f (x) = 7 + 6x − x 2 f ′(x) = 6 − 2x

The function is increasing when f ′(x) > 0 . First solve the related equation. 0 = 6 − 2x x=3 Test the inequality for arbitrary values in the intervals (–∞, 3) and (3, ∞). For the first interval, use x = 0. 4>0 L.S. = 6 − 2(0) =4

For the second interval, use x = 4. L.S. = 6 − 2(4) –2 < 0 = −2

f (x) is increasing on the interval (–∞, 3). f (x) is decreasing on the interval (3, ∞). b) Let y = f (x). f (x) = x 3 − 48x + 5 f ′(x) = 3x 2 − 48

The function is increasing when f ′(x) > 0 . First solve the related equation. 3 x 2 − 48 = 0 x 2 = 16 x = ±4

Test the inequality for arbitrary values in the intervals (–∞, –4), (–4, 4), and (4, ∞). For the first interval, use x = –5. L.S. = 3(−5)2 − 48 27 > 0 = 27

For the second interval, use x = 0. L.S. = 3(0)2 − 48 –48 < 0 = −48


368

For the third interval, use x = 5. 27 > 0 L.S. = 3(5)2 − 48 = 27

f (x) is increasing on the intervals (–∞, –4) and (4, ∞). f (x) is decreasing on the interval (–4, 4). c)

g(x) = x 4 − 18x 2 g ′(x) = 4x 3 − 36x

The function is increasing when g ′(x) > 0 . First solve the related equation. 0 = 4x 3 − 36x 0 = 4x(x 2 − 9) x = 0, x = ±3

Test the inequality for arbitrary values in the intervals (–∞, –3), (–3, 0), (0, 3), and (3, ∞). For the first interval, use x = –5. –320 < 0 L.S. = 4(−5)3 − 36(−5) = −320

For the second interval, use x = –1, 27 > 0 L.S. = 4(−1)3 − 36(−1) = 32

For the third interval, use x = 1. –32 < 0 L.S. = 4(1)3 − 36(1) = −32

For the fourth interval, use x = 5. L.S. = 4(5)3 − 36(5)

320 > 0

= 320

g(x) is increasing on the intervals (–3, 0) and (3, ∞). g(x) is decreasing on the intervals (–∞, –3), (0, 3). d)

f (x) = x 3 + 10x − 9 f ′(x) = 3x 2 + 10

The function is increasing when f ′(x) > 0 . First solve the related equation. 0 = 3x 2 + 10 10 x2 = − 3 There are no solutions to this equation. Test the inequality for one arbitrary value, such as x = 0.


369

L.S. = 3(0)2 + 10

10 > 0

= 10

f (x) is always increasing. Chapter 3 Review

Question 2 Page 204

f ′(x) = x(x − 3)2 0 = x(x − 3)2

x = 0, x = 3 Test points in the intervals. f ′(−1) = −16 f ′(1) = 4 f ′(4) = 4

x<0 Negative Decreasing

f ′(x) f (x)

Chapter 3 Review

x =0 0 Minimum

0<x<3 Positive Increasing

x=3 0 Point of inflection

x>3 Positive Increasing

Question 3 Page 204

a) Let y = f (x). f (x) = 3x 2 + 24x − 8 f ′(x) = 6x + 24 0 = 6x + 24 x = −4

Use a table to show increasing and decreasing intervals for the function.

Test value f ′(x)

y

x < –4

x = –4

x > –4

–5

–4

0

f ′(−5) = −6 Negative Decreasing

0

f ′(0) = 24 Positive Increasing

(–4, –56)

The critical point (–4, –56) is a local minimum. b)

f (x) = 16 − x 4 f ′(x) = −4x 3 0 = −4x 3 x=0


370

Use a table to show increasing and decreasing intervals for the function.

Test value f ′(x) f (x)

x<0

x=0

x>0

–1

0

1

f ′(−1) = 4 Positive Increasing

0

f ′(1) = −4 Negative Decreasing

(0, 16)

The critical point (0, 16) is a local maximum. c)

g(x) = x 3 + 9x 2 − 21x − 12 g ′(x) = 3x 2 + 18x − 21 0 = 3x 2 + 18x − 21 0 = 3(x + 7)(x − 1) x = −7, x = 1

Use a table to show increasing and decreasing intervals for the function. x < –7 x = –7 –7 < x < 1 x=1 Test –10 –7 0 1 value g ′(−10) = 99 g ′(0) = −21 g ′(x) 0 0 Positive Negative g(x) Increasing (–2, 233) Decreasing (1, –23)

x>1 2 g ′(2) = 27 Positive Increasing

The critical point (–2, 233) is a local maximum and the point (1, –23) is a local minimum. Chapter 3 Review

Question 4 Page 204

a) Find the critical points for this function. v(t) = 3t 2 − 24t + 88 v ′(t) = 6t − 24 6t − 24 = 0 t=4

Check that it is a minimum.


371

x<4 3 v ′(3) = −6 Negative Decreasing

Test value v ′(t) v(t)

x=4 4

x>4 5 v ′(5) = 6 Positive Increasing

0 (4, 40)

The critical point (4, 40) is a local minimum. The minimum sped of the car is 40 km/h. b) There are no critical points that can be maxima in this interval. The maximum must occur at an endpoint. v(2) = 52 v(5) = 43 The maximum speed of the car in the interval is 52 km/h.

Chapter 3 Review

Question 5 Page 204

Find the critical numbers. f (x) = x 3 − 8x 2 + 5x + 2 f ′(x) = 3x 2 − 16x + 5 0 = 3x 2 − 16x + 5 0 = (3x − 1)(x − 5) 1 x = ,x = 5 3

The critical numbers are

1 and 5. 3

To find local extrema, use a table to show increasing and decreasing intervals for the function. x<

1 3

x =

1 3

1 3

1 <x<5 3

x=5

x>5

1

5

6

Test value

0

f ′(x)

f ′(0) = 5 Positive

0

f ′(1) = −8 Negative

0

f ′(6) = 17 Positive

f (x)

Increasing

 1 76   ,   3 27 

Decreasing

(5, –48)

Increasing

 1 76  The critical point  , is a local maximum and the critical point (2, –48) is local minimum.  3 27 


372

Check the function values at the endpoints of the interval. f (0) = 2 f (6) = −40  1 76  The absolute minimum is (5, –48) and the absolute maximum is  ,  for the interval [ 0, 6] .  3 27 

Chapter 3 Review

Question 6 Page 204

B is the correct answer. A cubic function will have a quadratic (second degree) first derivative and a linear (first degree) second derivative. A linear function always has exactly one zero and thus a cubic function must have exactly one point of inflection. Chapter 3 Review

Question 7 Page 204

True. The second derivative of a quadratic function is of second degree. This derivative will have either zero, one, or two roots. Zero or two roots lead to zero or two points of inflection. If the quadratic has only one root, it means that it is always positive or always negative, which means that it does not change sign. In this case there can be no point of inflection because there is no sign change. Chapter 3 Review

Question 8 Page 204

f (x) = x 4 − 2x 3 − 12x 2 + 3 f ′(x) = 4x 3 − 6x 2 − 24x f ′′(x) = 12x 2 − 12x − 24

Find the possible points of inflection. 0 = 12x 2 − 12x − 24 0 = x2 − x − 2 0 = (x − 2)(x + 1) x = 2, x = −1

These values divide the domain into three intervals. Test the value of f ′′(x) in each interval and summarize in a table.

Test value f ′′(x) f (x)

x < –1

x = –1

–1 < x < 2

x=2

x>2

–2

–1

0

2

3

f ′′(−2) = 48 Positive

0

f ′′(0) = −24 Negative

0

f ′′(3) = 48 Positive

Concave up

Point of inflection (–1, –6)

Concave down

Point of Inflection (2, –45)

Concave up


373

Chapter 3 Review

Question 9 Page 204

a) From the graph of f ′(x) , f is increasing when −2 < x < 0 and x > 2 and deceasing otherwise. It will have local minima at x = ±2 and a local maximum at x = 0 . b) From the graph of f ′(x) , f ′′(x) will have zeros at –1.25 and +1.25. f ′′(x) will have negative values between –1.25 and +1.25 and positive values elsewhere. Since f ′(x) has an inflection point when x = 0 , f ′′(x) will have a local minimum at x = 0 .

Chapter 3 Review


Find the critical values. f (x) = 2x 3 − x 4 f ′(x) = 6x 2 − 4x 3 0 = 6x 2 − 4x 3 0 = 2x 2 (3 − 2x)

x = 0, x =

3 2

Use the second derivative test to determine maxima and minima. f ′′(x) = 12x − 12x 2 Test the critical points.  3 f ′′   = −9  2

() f ′′ (−0.5)= −9 f ′′ (0.5)= 3

f ′′ 0 = 0

The second derivative test indicates that there is a local maximum at x = 1.5 . Since the sign of the second derivative changes at x = 0 , there is a point of inflection there. MHR • Calculus and Vectors 12 Solutions

374

To sketch the curve, more information is needed. Use a table to show increasing and decreasing intervals for the function.


x<0

x =0

0 < x < 1.5

x = 1.5

x > 1.5

–1

0

1

0

2

f ′(−1) = 10 Positive Increasing

0

f ′(1) = 2 Positive Increasing

0

f ′(2) = −8 Negative Decreasing

(0,0)

(1.5, 1.7)

Then sketch the graph.

Chapter 3 Review


Vertical asymptotes only occur where the denominator in a rational function equals zero. a) x = 0 b) 0 = 2x − 4 x=2 c) 0 = x 2 − 3x − 10 0 = (x − 5)(x + 2) x = 5, x = −2

d) 0 = x 2 + 2x + 1 0 = (x + 1)(x + 1) x = −1

Chapter 3 Review


a) Substituting a small, positive number for x means the numerator is close to 4 and the denominator is a very small, but positive number. The quotient of these two numbers will make a very large, positive number. Thus, lim f ( x ) = ∞ . x →0+


375

b) The limit has the same value, lim f ( x ) = ∞ , since substituting a small but negative number will still x → 0−

lead to a quotient of 4 and a small positive number. c) Let y = 0 . x+4 =0 x2 x+4=0 x = −4

d) To find the turning point, let the derivative equal zero. f (x) = (x + 4)x −2 f ′(x) = (x + 4)(−2)x −3 + (1)x −2 = x −3 (−2x − 8 + x) 0 = x −3 (−2x − 8 + x) 0 = −x − 8 x = −8 1  The turning point is  −8, −  16  

e)

Chapter 3 Review


a) f (−x) = (−x)3 − 3(−x) = −x 3 + 3x = − f (x) Therefore the function is an odd function.

b) This is a polynomial function. The domain is . c) Let x = 0. f (0) = (0)3 − 3(0) =0 The y–intercept is 0.


376

Let y = 0. 0 = x 3 − 3x 0 = x(x 2 − 3) x = 0, x = ± 3

There are three x-intercepts at – 3 , 0, and

3.

d) Find the critical points. f (x) = x 3 − 3x f ′(x) = 3x 2 − 3 f ′′(x) = 6x 0 = 3x 2 − 3 0 = 3(x 2 − 1) x = ±1

Use the second derivative test for extrema. f ′′(−1) = 6(−1) = −6 f ′′(1) = 6(1) =6

Therefore (1, –2) is a local minimum and (–1, 2) is a local maximum. The critical points create three intervals. Test the derivative in each interval.


x < –1 –2 f ′(−2) = 9 Positive Increasing

–1 < x < 1 0 f ′(0) = −3 Negative Decreasing

x>1 2 f ′(2) = 9 Positive Increasing

The function is increasing for x < −1 and x > 1 and is decreasing for −1 < x < 1 . For concavity, need to find all points of inflection. Let f ′′(x) = 0 . 6x = 0 x=0


377

Test for concavity in the intervals between the possible points of inflection. x<0 Test value

x =0

–1

x >0 1

f ′′(x)

f ′′(−1) = −6 Negative

0

f ′′(1) = 6 Positive

f (x)

Concave down

Point of Inflection (0, 0)

Concave up

Chapter 3 Review


a) Follow the six-step plan. f (x) = −x 2 + 2x f ′(x) = −2x + 2 f ′′(x) = −2

Step 1: Since this is a polynomial function, the domain is  . Step 2: f (0) = 0 . The y-intercept is 0. 0 = −x 2 + 2x 0 = x(x − 2) x = 0, x = 2 The x-intercepts are 0 and 2.

Step 3: Find the critical numbers. 0 = −2x + 2 x =1 Use the second derivative test to classify the critical points. f ′′(1) = −2 Therefore (1, 1) is a local maximum. Step 4: Find the possible points of inflection. Since f ′′(x) − 2 ≠ 0 for any x, there are no points of inflection. Step 5: From Step 3, f (x) is decreasing for x > 1 and increasing for x < 1 . From Step 4, f (x) is concave down for all x.


378

Step 6: Sketch the graph.

b) Follow the six-step plan. 1 9 k(x) = x 4 − x 2 4 2 3 k ′(x) = x − 9x k ′′(x) = 3x 2 − 9

Step 1: Since this is a polynomial function, the domain is  . Step 2: k(0) = 0 . The y-intercept is 0. 1 4 9 2 x − x 4 2 2 2 0 = x (x − 18) 0=

x = 0, x = ± 18

The x-intercepts are 0 and ±3 2 . Step 3: Find the critical numbers. 0 = x 3 − 9x 0 = x(x 2 − 9) x = 0, x = ±3

Use the second derivative test to classify the critical points. k ′′(−3) = 18 k ′′(0) = −9 k ′′(3) = 18

Therefore (–3, –20.25) and (3, –20.25) are local minimum points and (0, 0) is a local maximum.


379

Step 4: Find the possible points of inflection. 0 = 3x 2 − 9 x2 = 3 x=± 3

Test the intervals. Have already tested x= 0, − 3, and 3 .

(

) ( 6, − 18)are points of inflection.

Therefore − 6, − 18 and

Step 5: From Step 3, k(x) is decreasing for x < −3 and 0 < x < 3 and increasing for −3 < x < 0 and x > 3 . From Step 4, k(x) is concave down for − 6 < x < 6 and concave up elsewhere. Step 6: Sketch the graph.

c) Follow the six-step plan. h(x) = 2x 3 − 3x 2 − 3x + 2 h′(x) = 6x 2 − 6x − 3 h′′(x) = 12x − 6

Step 1: Since this is a polynomial function, the domain is  . Step 2: h(0) = 2. The y-intercept is 2. 0 = 2x 3 − 3x 2 − 3x + 2 0 = 2(x 3 + 1) − 3x(x + 1) 0 = 2(x + 1)(x 2 − x + 1) − 3x(x + 1) 0 = (x + 1)(2x 2 − 2x + 2 − 3x) 0 = (x + 1)(2x 2 − 5x + 2) 0 = (x + 1)(2x − 1)(x − 2) x = −1, x =

1 , x=2 2


380

The x-intercepts are –1,

1 and 2. 2

Step 3: Find the critical numbers. 0 = 6x 2 − 6x − 3 0 = 3(2x 2 − 2x − 1) x=

2 ± 12 4

1± 3 2 x B −0.37, x B1.37  x=

Use the second derivative test to classify the critical points. h′′(−0.37) = −10.44 h′′(1.37) = 10.44

Therefore (–0.37, 2.60) is a local maximum point and (1.37, –2.60) is a local minimum. Step 4: Find the possible points of inflection. 0 = 12x − 6 1 x= 2 Have already tested the intervals around 0.5. Therefore (0.5, 0) is a point of inflection. Step 5: From , h(x) is decreasing for

1− 3 1+ 3 and increasing <x< 2 2

1− 3 1+ 3 . and x > 2 2 From , h(x) is concave down for x < 0 and concave up for x > 0 .

for x <


381


QuickTime™ and a TIFF (LZW) decompressor are needed to see this picture.

Chapter 3 Review


a) Find the critical points. 0.12t C(t) = 2 t + 2t + 2 = 0.12t(t 2 + 2t + 2)−1 C ′(t) = 0.12t(−1)(t 2 + 2t + 2)−2 (2t + 2) + (0.12)(t 2 + 2t + 2)−1 = (t 2 + 2t + 2)−2 (−0.24t 2 − 0.24t + 0.12t 2 + 0.24t + 0.24) =

−0.12t 2 + 0.24 (t 2 + 2t + 2)2

0 = −0.12t 2 + 0.24 0 = −0.12(t 2 − 2) t=± 2

Since there is only one critical value in the interval 0 ≤ x ≤ 4 , test the function at the critical point and the endpoints. C(0) = 0 C

( 2 )= 0.025 C(4) = 0.018

The maximum concentration of the drug during the interval 0 ≤ x ≤ 4 is 0.025 mg/cm3. b) The maximum concentration occurs approximately 1.414 hours after istration of the drug.


382

Chapter 3 Review


Use the variables suggested in the question. Let S.A. be the surface area, the variable to be minimized. S.A. = 1(2w2 ) + 2(2wh) + 2(wh) = 4w2 + 6wh

Volume is to be 2400 cm2. V = 2w2 h 2400 = 2w2 h h=

1200 w2

Simplify the surface area function.  1200  S.A. = 2w2 + 6w  2   w  = 2w2 + 7200w−1

For a minimum, find the critical points. S.A.′ = 4w − 7200w−2 0 = 4w − 7200w−2 0 = 4w3 − 7200 w3 = 1800 w = 3 1800  w B12.16

Test values on either side of this point. S.A.′ (11) = −15.5 S.A.′ (13) = 9.40

Since the slope changes sign, the point must be a minimum. 1200 2w = 24.32 h= (12.16)2 w = 12.16 = 8.11 The dimensions of the box will be 12.16 cm by 24.32 cm by 8.11 cm.


383

Chapter 3 Practice Test Chapter 3 Practice Test

Question 1 Page 206

B is the correct answer. f ′(x) = 2x − 8 0 = 2x − 8 x=4 f ′(0) = −8 f ′(3) = −2 The only critical point is not in the given interval. The function is decreasing at both endpoints of the interval. Therefore it is always decreasing on the interval.

Chapter 3 Practice Test

Question 2 Page 206

D is the correct answer. A, B, and C are incorrect. The function is decreasing for all values of x. Chapter 3 Practice Test

Question 3 Page 206

C is the correct answer. A: (2, f (2)) is a critical point because f ′(2) = 0 . B: (2, f (2)) (2, f (2)) might be a turning point. D: (2, f (2)) (2, f (2)) could be a local maximum (or a point of inflection) since f is increasing on the interval preceding x = 2. It certainly cannot be a local minimum. Chapter 3 Practice Test

Question 4 Page 206

C is the correct answer. For an odd function, f (−x) = − f (x) . Therefore: f (−a) = − f (a) = −5


384


Question 5 Page 206

C is the correct answer. A: The graph has no x-intercepts since the numerator of the fraction cannot be zero. B: f (x) = −3(x − 2)−2 f ′(x) = 6(x − 2)−3 f ′′(x) = −18(x − 2)−4 −18 (x − 2)4 The second derivative is always negative (the denominator is always positive). The curve is always concave down. 6 C: f ′(0) = −8 = −0.75 This contradicts the statement f ′(x) > 0 when x < 2 . D: If a number close to 2 is substituted for x, the fraction becomes –3 divided by a very small positive number, the result of which is a very large negative number. =


Question 6 Page 206

Given f ′(x) = x(x − 1)2 , the graph of f (x) has two critical points and one turning point. f ′(x) = x(x − 1)2 0 = x(x − 1)2 x = 0, x = 1

Check the intervals. f ′(−1) = −4 f ′(0.5) = 0.125 f ′(2) = 2

There is a minimum when x = 0 and a point of inflection when x = 1. Chapter 3 Practice Test

Question 7 Page 206

A corresponds to a local maximum since slopes for f (x) will change from positive to negative at that point. B corresponds to a point of inflection since slopes for f (x) will not change at that point but will be negative before and after the point. C corresponds to a local minimum since slopes for f (x) will change from negative to positive at that point.


385


Question 8 Page 206

f (x) = x 3 − 5x 2 + 6x + 2 f ′(x) = 3x 2 − 10x + 6 0 = 3x 2 − 10x + 6 x=

10 ± 28 6

5± 7 3 x B 2.5, x B 0.8  x=

Check values at the critical points and the endpoints of the interval. f (0) = 2  5− 7  f  B 4.1  3   5+ 7  f  B1.4  3  

f (4) = 10

The absolute maximum is (4, 10) and the absolute minimum is (2.5, 1.4) Chapter 3 Practice Test

Question 9 Page 206

Answers may vary. For example: f (x) in red, f ′(x) in blue, and f ′′(x) in green. Note that zeros of f ′(x) correspond to local extrema on f (x) and zeros of f ′′(x) correspond to points of inflection on f (x) . 60 40 20

-1

1

2

3

-20 -40 -60


386



a) This is a quadratic polynomial function with leading coefficient +3. Would expect the limit to be +∞ in both cases. Substitute large numbers for x to check. f (100) = 284 180 000 f (−100) = 316 018 000

Clearly lim f ( x ) = ∞ and lim f ( x ) = ∞. x →−∞

b)

x →∞

f (x) = 3x 4 − 16x 3 + 18x 2 f ′(x) = 12x 3 − 48x 2 + 36x f ′′(x) = 36x 2 − 96x + 36

Find the critical points. 0 = 12x 3 − 48x 2 + 36x 0 = 12x(x 2 − 4x + 3) 0 = 12x(x − 3)(x − 1) x = 0, x = 1, x = 3

Test the critical points using the second derivative test. f ′′(0) = 36(0)2 − 96(0) + 36 = 36 f ′′(1) = 36(1)2 − 96(1) + 36 = −24 f ′′(3) = 36(3)2 − 96(3) + 36 = 72

There are local minima at (0, 0) and (3, –27) and a local maximum at (1, 5). c) For possible points of inflection, set the second derivative to zero. 0 = 36x 2 − 96x + 36

(

)

0 = 12 3x 2 − 8x + 3 x=

8 ± 64 − 36 6

4± 7 3 x B 0.45, x B 2.22  x=

Test values for the intervals created by these points have already been checked above. There are two points of inflection at (0.45, 2.31) and (2.22, –13.48).


387



a) C(x) = 0.1x 2 + 1.2x + 3.6 0.1x 2 + 1.2x + 3.6 x = 0.1x + 1.2 + 3.6x −1

U (x) =

b) To minimize unit cost, find the critical points for U(x). U ′(x) = 0.1− 3.6x −2 0 = 0.1− 3.6x −2 0 = x 2 − 36 x = ±6

The negative value has no meaning in this situation. Check the values around x = 6 . U ′(5) = 0.1− 3.6(5)−2 = −0.044 U ′(7) = 0.1− 3.6(7)−2 = 0.027

There is a minimum when x = 6. The company should produce 6 ATVs per day to minimize the cost per unit. Chapter 3 Practice Test


a) Vertical asymptotes occur when the denominator in a rational function is zero. In this case, the vertical asymptote is x = 0.

b) Let y = f(x). 1 x2 = x 2 + x −2

f (x) = x 2 +

f ′(x) = 2x − 2x −3 f ′′(x) = 2 + 6x −4

Find the critical points. 0 = 2x − 2x −3 0 = x4 − 1 x = ±1


388

Test using the second derivative test. f ′′(−1) = 8 f ′′(1) = 8

There are local minima at (1, 2) and (–1, 2). c) To find intervals of concavity, let f ′′(x) = 0 0 = 2 + 6x −4 0 = 2x 4 + 6 x 4 = −3

There are no real roots to this equation. The only value dividing regions of concavity is related to the asymptote. Based on earlier tests, the function is concave up for x < 0 and also x > 0 . d)

Chapter 3 Practice Test a) i)


f ′(0) B −3 f ′(1) B −4 f ′(0) > f ′(1) 

ii) Between x = –1 and x = 3, all the slopes for f are negative (below zero on the f ′(x) graph). Therefore, the function is decreasing from x = –1 to x = 3 and f (−1) > f (3) . iii) There are no critical numbers between 5 and 10 and all the slopes of f are positive in this interval. Therefore, f (10) > f (5)


389

b) The sketch should have a maximum at x = –1 and, a minimum x = 3, and a point of inflection at x = 1.



Answers may vary. For example: Since the factor in the denominator is squared, all of the values of the function will be positive. The curve will always be above the x-axis. Therefore, the limit as an asymptote is approached from either side must be +∞ Chapter 3 Practice Test


a) Let x be the number of $10 reductions in price. The revenue function is the number of rooms rented times the price per room. R(x) = (40 + 10x)(120 − 10x) = −100x 2 + 800x + 4800 R′(x) = −200x + 800 0 = −200x + 800 x=4 To maximize revenue, the hotel should decrease the price by four increments to $80 per room.

b) The revenue function is increasing until x = 4. If there are only 50 rooms, the hotel can only make one decrease of $10. Compare the revenues. R(0) = 4800 R(1) = 5500 R(4) = 6400

With 50 rooms, the maximum revenue is $5500.


390



a)

1 − 2x 2

x Let the perimeter be 1 unit. The dimensions of the rectangle will be x and

1 − 2x . 2

 1− 2x  A(x) = x   2  = 0.5x − x 2

For a maximum area: A′(x) = 0.5 − 2x 0 = 0.5 − 2x x = 0.25 1 − 2 ( 0.25 ) 2

= 0.25

Therefore the dimensions are 0.25 units by 0.25 units. The maximum area rectangle is a square. b)

x

x h

1 − 2x 2

Assume that the triangle is isosceles.


391

Then for perimeter = 1,  1− 2x  x 2 = h2 +   2  x 2 = h2 +

2

1 − x + x2 4

1 h= − +x 4  1 A(x) =   1− 2x  2

(

)

1 − +x 4 1

= (0.5 − x)(x − 0.25) 2 1 1 −  1 A′(x) = (Ğ1)(x − 0.25) 2 + (0.5 − x)   (x − 0.25) 2  2

For a maximum, let A′ ( x ) = 0 . 1 1 −  1 (Ğ1)(x − 0.25) 2 + 0.5 − x   (x − 0.25) 2 = 0  2

(

)

1

−(x − 0.25) + (0.5 − x)0.5 = 0

Multiply by (x − 0.25) 2 .

0.25 − x + 0.25 − 0.5x = 0 1 x= 3 1 Therefore each side must be of the perimeter. 3

c) For a given perimeter, the octagon will enclose more area than the pentagon. The circle is the most efficient shape for a given perimeter in two dimensions and the octagon is closer to the circle shape than the pentagon would be. Again, the proof of these statements is beyond the scope of this course, but an investigation with specific numbers could confirm them. d) In general, a circle or a square will enclose a maximum area for a perimeter. Chapter 3 Practice Test


a) For a maximum yield, let the derivative equal zero. B(n) = −0.1n2 + 10n B′(n) = −0.2n + 10 0 = −0.2n + 10 n = 50

They should plant 50 000 seeds per acre for a maximum yield.


392

b) Create a profit function. P(n) = 3(−0.1n2 + 10n) − 2n = −0.3n2 + 28n

For a maximum profit, let the derivative equal zero. P′(n) = −0.6n + 28 0 = −0.6n + 28 n = 46.667

For maximum profit, they should plant about 47 000 seeds per acre. Chapter 3 Practice Test


x

x h

32 − x

Let the two equal sides of the isosceles triangle have length x cm. In the right triangle: 2 x 2 =h 2 + ( 32 − x ) h =−1024 + 64 x

The area of the isosceles triangle is:  1 A(x) =   (64 − 2x) 64x − 1024  2 = (32 − x) 64x − 1024


393

For a maximum: 1 −  1 A′(x) = (−1) 64x − 1024 + (32 − x)   (64x − 1024) 2 (64)  2

A′(x) = − 64x − 1024 + (1024 − 32x)(64x − 1024) 0 = − 64x − 1024 + (1024 − 32x)(64x − 1024)

−

1 2

−

1 2

0 = −64x + 1024 + 1024 − 32x 96x = 2048 64 3  x B 21.33 x=

For a maximum area the three sides need to have the same length of approximately 21.33 cm.


394

Chapter 1 to 3 Review Chapter 1 to 3 Review a)

Question 1 Page 208

h(20) − h(10) 0.04 − 0.82 = 20 − 10 10 =&−0.078 

The average rate of change of the height is –0.078 m/s. This is the average vertical velocity between 10 s and 20 s. b)

h(20 + 0.1) − f (20) 0.02 917 − 0.04 = 0.1 0.1 =&−0.11 

The instantaneous rate of change at t = 20 s is approximately –0.11 m/s. This is the instantaneous vertical velocity at t = 20 s. c)

Chapter 1 to 3 Review

Question 2 Page 208

15 12 9 6 3 a) t1 = ;Êt2 = ;Êt3 = ;Êt4 = ;Êt5 = 26 17 10 5 2

b) Yes, the limit is zero. As x increases, the denominator increases faster than the numerator so the sequence approaches 0. Chapter 1 to 3 Review

Question 3 Page 208

The limits as x approaches 5 from the left and right sides are both equal to –2, so the limit as x approaches 5 exists and equals –2. The value of the function at x = 5 is not –2, so there is a removable discontinuity at this point. The function is discontinuous at x = 5.


395


Question 4 Page 208

a) lim (−2x 3 + x 2 − 4x + 7) = −50 x→3

x 2 − 16 (x − 4)(x + 4) b) lim = lim x→4 x − 4 x→4 x−4 = lim (x + 4) x→4

=8

5 3x − 5 =− 3 x→0 7 + 4x 7

c) lim


Question 5 Page 208

a)  {x | x ≠ 3, x ∈° } b) lim+ f (x) = ∞ x→3

c) No. The graph is discontinuous at x = 3, since there is a vertical asymptote at this point. Chapter 1 to 3 Review

Question 6 Page 208

a)

b)


396


a)

dy = lim dx h→0

Question 7 Page 208

2  2 (x + h)3 − 8(x + h) −  x 3 − 8x  3 3  h

2 3 2 (x + 3x 2 h + 3xh2 + h3 ) − 8(x + h) − x 3 + 8x 3 = lim 3 h→0 h 2 2x 2 h + 2xh2 + h3 − 8h 3 = lim h→0 h 2 = lim (2x 2 + 2xh + h2 − 8) h→0 3 2 = 2x − 8

b)

2(x + h) − 1 − 2x − 1 dy = lim dx h→0 h = lim h→0

( 2(x + h) − 1 − 2x − 1)( 2(x + h) − 1 + h ( 2(x + h) − 1 + 2x − 1 ) 2(x + h) − 1− (2x − 1)

= lim h→0

h

c)

)

2(x + h) − 1 + 2x − 1

h→0

=

2x − 1

)

2

= lim =

( 2(x + h) − 1 +

2x − 1

2 2 2x − 1 1 2x − 1

dy 3(x + h)2 − (x + h) + 1− (3x 2 − x + 1) = lim h dx h→0 2 2 3x + 6xh + 3h − x − h + 1− 3x 2 + x − 1 = lim h→0 h 2 6xh + 3h − h = lim h→0 h = lim (6x + 3h − 1) h→0

= 6x − 1


397

d) f ′(t) = lim

−2(t + h)  −2t  − (t + h) + 3  t + 3 

h (t + 3)(−2t − 2h) + 2t(t + h + 3) (t + h + 3)(t + 3) = lim h→0 h −6h = lim h→0 h(t + h + 3)(t + 3) h→0

= lim h→0

=−

−6 (t + h + 3)(t + 3)

6 (t + 3)2

Chapter 1 to 3 Review a)

b)

Question 8 Page 208

dy = 24x3 – 15x2 + 3 dx

dy = 1(4t − 7) + (3 + t)(4) dx = 8t + 5

c) g ′(x) = 4(−2x 3 + 3)3 (−6x 2 ) = −24x 2 (−2x 3 + 3)3 1 1 −  1 (5t − 1) 2 (1) − t   (5t − 1) 2 (5)  2 d) s′(t) = 5t − 1 1 5t = − 1 3

(5t − 1) 2

=

2(5t − 1) 2

2(5t − 1) − 5t 3

2(5t − 1) 2 =

5t − 2 3

2(5t − 1) 2


398

Chapter 1 to 3 Review a)

Question 9 Page 208

dy (x + 3)2 (−4) + 4x(2)(x + 3) = dx (x + 3)4 =

−4x 2 − 24x − 36 + 8x 2 + 24x (x + 3)4

=

4x 2 − 36 (x + 3)4

=

4(x + 3)(x − 3) (x + 3)4

=

4(x − 3) (x + 3)3

The slope of the tangent at x = –2 is dy dx

= x=−2

dy dx

x=−2

4(−2 − 3) (−2 + 3)3

= −20

When x = –2, y = 8. Use the point (–2, 8) and m = –20 to find b in the equation of the tangent, y = mx + b. 8 = –20(–2) + b b = –32 The equation of the tangent at x = –2 is y = –20x – 32. b) The slope of the tangent at x = –1 is dy dx

= x=−1

dy dx

x=−1

4(−1− 3) (−1+ 3)3

= −2  1 1 The normal is perpendicular to the tangent, so the slope of the normal is −   = .  −2  2


399

When x = –1, y = 1. Use the point (–1, 1) and m =

1 to find b in the equation of the normal, y = mx + b. 2

1 (–1) + b 2 3 b= 2

1=

The equation of the normal at x = –1 is y = Chapter 1 to 3 Review

1 3 x+ . 2 2


a) y – x + 2 = 0 y=x–2 The slope of the tangent lines is perpendicular to the line above, so the tangents have a slope of –1. Now, find f ′(x) and the values of x where f ′(x) = −1 . f ′(x) = 3x 2 + 4x −1 = 3x 2 + 4x 0 = 3x 2 + 4x + 1 0 = (3x + 1)(x + 1) 1 x = − ,Êx = −1 3

 1 5 Therefore, the points are  − ,Ê  and (–1, 1).  3 27 

b) The slope of the tangents is –1.  1 5 Use the point  − ,Ê  and m = –1 to find b in the equation of the first tangent, y = mx + b.  3 27   1 5 = −−  + b 27  3 b=−

4 27

Use the point (–1, 1) and m = –1 to find b in the equation of the second tangent, y = mx + b. 1 = –1(–1) + b b=0


400

Therefore, the equation of the first tangent is y = −x −

4 and the equation of the second tangent is 27

y = –x. c) Sketch the graph of the function and the tangent lines.



dy dy du ⋅ = dx du dx 1  1 − = (5 − 4u + 3u )  (2 − x) 2 (−1)  2 2

( (

)

 −1  = 5 − 4 2 − x + 3(2 − x)   2 2 − x   −1  = 11− 4 2 − x − 3x   2 2 − x 

dy dx

)

  −1 = 11− 4 2 − (−2) − 3(−2)    2 2 − (−2) 

(

x=−2

)

= 2.25



a) i) Graph 1; Velocity and acceleration are derivatives of position. ii) Graph 2; Velocity is the derivative of position, which is cubic, so the velocity function is quadratic.


401

iii) Graph 3; Acceleration is the derivative of velocity, which is quadratic, so the acceleration function is linear. b) The object is slowing down when v(t) × a(t) < 0 , so over the intervals x < 1 and 2 < x < 3. The object is speeding up when v(t) × a(t) > 0 , so over the intervals 1 < x < 2 and x > 3. c) If the slope of the position function is moving towards zero then the object is slowing down. If the slope is moving away from zero then the object is speeding up. Chapter 1 to 3 Review


a) V (t) = (3.72 − 0.02t 2 )(13.00 + 0.21t) = −0.0042t 3 − 0.26t 2 + 0.7812t + 48.36

b) V ′(t ) = –0.0126t2 – 0.52t + 0.7812 This is the change of voltage as time changes. c) i) V ′(3) = Ğ0.0126(3)2 Ğ0.52(3) + 0.7812 = −0.8922

The rate of change of voltage at t = 3 s is –0.8922 V/s. ii) I ′(t) = −0.04t I ′(3) = −0.04(3) = −0.12

The rate of change of current at t = 3 s is –0.12 A/s. iii) R′(t) = 0.21 R′(3) = 0.21

The rate of change of resistance at t = 3 s is 0.21 Ω/s. Chapter 1 to 3 Review


a) C ′(x) = −0.002x + 1.5 C ′(300) = 0.90

The marginal cost is $0.90 per gadget. b) C(301) − C(300) = −0.001(301)2 + 1.5(301) + 500 − (−0.001(300)2 + 1.5(300) + 500) = 0.899

The actual cost of producing the 301st gadget is $0.90.


402

c)

R(x) = xp(x) = 4.5x − 0.1x 2 R′(x) = 4.5 − 0.2x

R′(300) = 4.5 − 0.2(300) = −55.50

The marginal revenue is –$55.50 per gadget. P(x) = R(x) − C(x) = −0.099x 2 + 3x − 500 P′(x) = −0.198x + 3 P′(300) = 3 − 0.198(300)

= −56.40

The marginal profit is –$56.40 per gadget. Chapter 1 to 3 Review a) i)


f ′(2) since f ′(−2) = 6ÊandÊf ′(2) = 10 .

ii) f (6) since f decreases from x = 6 to x = 12. b)


403



a) f ′(x) = 4x 3 − 12x 2 0 = 4x 2 (x − 3) x = 0,Êx = 3

When x < 0, f ′(x) < 0 . When 0 < x < 3, f ′(x) < 0 . When x > 3, f ′(x) > 0 . Therefore, (0, 0) is a point of inflection and (3, –27) is a local minimum. b)

dy = 6x 2 − 6x − 11 dx 0 = 6x 2 − 6x − 11 −(−6) ± (−6)2 − 4(6)(−11) x= 2(6) 3±5 3 6 x =&1.94,Êx =&−0.94 x=

When x < –0.94, f ′(x) > 0 . When –0.94 < x < 1.94, f ′(x) < 0 . When x > 1.94, f ′(x) > 0 . Therefore, (–0.94, 12) is a local maximum and (1.94, –12) is a local minimum. c) h′(x) = 4x 3 − 15x 2 + 2x + 21 0 = (x − 3)(4x 2 − 3x − 7) 0 = (x − 3)(4x − 7)(x + 1) x = −1,Êx =

7 ,Êx = 3 4

When x < –1, f ′(x) < 0 . When –1 < x <

7 , f ′(x) > 0 . 4

7 < x < 3, f ′(x) < 0 . 4 When x > 3, f ′(x) > 0 .

When

 7 1125  Therefore, (–1, –32) is a local minimum,  ,Ê is a local maximum, and (3, 0) is a local  4 256  minimum.


404

d) g ′(x) = 12x 3 − 24x 2 − 12x + 24 0 = (x − 1)(12x 2 − 12x − 24) 0 = 12(x − 1)(x − 2)(x + 1) x = −1,Êx = 1,Êx = 2

When x < –1, f ′(x) < 0 . When –1 < x < 1, f ′(x) > 0 . When 1 < x < 2, f ′(x) < 0 . When x > 2, f ′(x) > 0 . Therefore, (–1, –28) is a local minimum, (1, 4) is a local maximum, and (2, –1) is a local minimum. Chapter 1 to 3 Review


a) Follow the six-step plan. f (x) = x 3 − 9x 2 + 15x + 4 f ′(x) = 3x 2 − 18x + 15 f ′′(x) = 6x − 18

Step 1:Since this is a polynomial function, the domain is  . Step 2: f (0) = 4 . The y-intercept is 4. Since x 3 − 9 x 2 + 15 x + 4 = 0 is not factorable, do not look for x-intercepts at this time. Step 3: Find the critical numbers. 3x 2 − 18x + 15 = 0 3(x 2 − 6x + 5) = 0 3(x − 5)(x − 1) = 0 x = 5, x = 1

Use the second derivative test to classify the critical points. f ′′(1) = −12 f ′′(5) = 12

Therefore (5, –21) is a local minimum point and (1, 11) is a local maximum. Step 4: Find the possible points of inflection. 6 x − 18 = 0 x=3 Have already tested the intervals around x = 3. Therefore (3, –5) is a point of inflection. Step 5: From Step 3, f is decreasing for 1 < x < 5 and increasing for x < 1 and x > 5 . From Step 4, f is concave down for x < 3 and concave up for x > 3 . MHR • Calculus and Vectors 12 Solutions

405


b) Follow the six-step plan. f (x) = x 3 − 12x 2 + 36x + 5 f ′(x) = 3x 2 − 24x + 36 f ′′(x) = 6x − 24

Step 1: Since this is a polynomial function, the domain is  . Step 2: f (0) = 5 . The y-intercept is 5. Since x 3 − 12 x 2 + 36 x + 5 = 0 is not factorable, do not look for x-intercepts at this time. Step 3: Find the critical numbers. 3x 2 − 24x + 36 = 0 3(x 2 − 8x + 12) = 0 3(x − 6)(x − 2) = 0 x = 2, x = 6

Use the second derivative test to classify the critical points. f ′′(2) = −12 f ′′(6) = 12

Therefore (6, 5) is a local minimum point and (2, 37) is a local maximum.  Find the possible points of inflection. 6 x − 24 = 0 x=4 Have already tested the intervals around x = 4 . Therefore (4, 21) is a point of inflection.  From , f is decreasing for 2 < x < 6 and increasing for x < 2 and x > 6 . From , f is concave down for x < 4 and concave up for x > 4 .


406

 Sketch the graph.

c) Follow the six-step plan. 2x f (x) = 2 x − 5x + 4 = 2x(x 2 − 5x + 4)−1 f ′(x) = 2(x 2 − 5x + 4)−1 + 2x(−1)(x 2 − 5x + 4)−2 (2x − 5) = (x 2 − 5x + 4)−2 (2x 2 − 10x + 8 − 4x 2 + 10x) = (−2x 2 + 8)(x 2 − 5x + 4)−2 f ′′(x) = (−4x)(x 2 − 5x + 4)−2 + (−2x 2 + 8)(−2)(x 2 − 5x + 4)−3 (2x − 5) = (x 2 − 5x + 4)−3 (−4x 3 + 20x 2 − 16x + 8x 3 − 20x 2 − 32x + 80) = (4x 3 − 48x + 80)(x 2 − 5x + 4)−3

 Since this is a rational function, the domain is   except where there are vertical asymptotes. 2 x − 5x + 4 = 0 (x − 4)(x − 1) = 0 x = 4, x = 1 There are vertical asymptotes at x 1= = and x 4 .  f (0) = 0 . The y-intercept is 0. When= y 0,= x 0 is the only solution. The x-intercept is 0.  Find the critical numbers. 0 = (−2x 2 + 8)(x 2 − 5x + 4)−2 0 = −2x 2 + 8 x2 = 4 x = 2, x = –2

Use the second derivative test to classify the critical points. f ′′(2) = −2 f ′′(−2) = 0.003 Therefore (–2, –0.22) is a local minimum point and (2, –2) is a local maximum.


407

 Find the possible points of inflection. 4x 3 − 48x + 80(x 2 − 5x + 4)−3 = 0 4x 3 − 48x + 80 = 0 4(x 3 − 12x + 20) = 0

This cubic equation does have a root but it is not readily factorable. There will be at least one point of inflection.  From , f is decreasing for 2 < x < 4 and increasing for 1 < x < 2 . From , f is concave down for 1 < x < 4 .  Sketch the graph.

d) Follow the six-step plan. f (x) = 3x 3 + 7x 2 + 3x − 1 f ′(x) = 9x 2 + 14x + 3 f ′′(x) = 18x + 14

 Since this is a polynomial function, the domain is  .  f (0) = −1 . The y-intercept is –1. For x-intercepts, let y = 0 . 3x 3 + 7x 2 + 3x − 1 = 0

Try factoring using the factor theorem. f (1) = 12 f (–1) = 0 Therefore, (x + 1) is a factor. (x + 1)(3x 2 + 4x − 1) = 0 x = –1 and x =

−4 ± 28 or x  =–1.5, x  =0.2 6

There are x-intercepts at –1.5, –1, and 0.2.


408

 Find the critical numbers. 0 = 9x 2 + 14x + 3 −14 ± 88 18 x =–1.3, x  =0.3 x=

Use the second derivative test to classify the critical points. f ′′(−0.3) = 8.6 f ′′(−1.3) = −9.4

Therefore (–0.3, –1.4) is a local minimum point and (–1.3, 0.4) is a local maximum.  Find the possible points of inflection. 18 x + 14 = 0 7 x= − 9 7 Have already tested the intervals around x = − . 9 Therefore (–0.8, –0.5) is a point of inflection.

 From , f is decreasing for −1.3 < x < −0.3 and increasing for x < −1.3 and x > −0.3 . 7 7 From , f is concave down for x < − and concave up for x > − . 9 9  Sketch the graph.

e) Follow the six-step plan. f (x) = x 4 − 5x 3 + x 2 + 21x − 18 f ′(x) = 4x 3 − 15x 2 + 2x + 21 f ′′(x) = 12x 2 − 30x + 2

 Since this is a polynomial function, the domain is  .  f (0) = −18 . The y-intercept is –18. For x-intercepts, let y = 0 . x 4 − 5x 3 + x 2 + 21x − 18 = 0


409

Try factoring using the factor theorem. f (1) = 0 Therefore, (x – 1) is a factor. (x − 1)(x 3 − 4x 2 − 3x + 18) = 0 Try factoring again using the factor theorem. f (3) = 0 Therefore, (x – 3) is a factor. (x − 1)(x − 3)(x 2 − x − 6) = 0 (x − 1)(x − 3)(x − 3)(x + 2) = 0 x = 1, x = 3, x = –2

There are x-intercepts at –2, 1, and 3.  Find the critical numbers. 4x 3 − 15x 2 + 2x + 21 = 0 Use the factor theorem. f (–1) = 0 Therefore, (x + 1) is a factor. (x + 1)(4x 2 − 19x + 21) = 0 (x + 1)(x − 3)(4x − 7) = 0 x = –1, x = 3, x = 1.75

Use the second derivative test to classify the critical points. f ′′(−1) = 44 f ′′(3) = 20 f ′′(1.75) = −13.75

Therefore (–1, –32) and (3, 0) are local minimum points and (1.75, 4.4) is a local maximum.  Find the possible points of inflection. 0 = 12x 2 − 30x + 2 0 = 6x 2 − 15x + 1 15 ± 201 12 x =0.1, x  =2.4 x=

Have already tested the intervals around these values. Therefore (0.1, –16.6) and (2.4, 2.1) are points of inflection.  From , f is decreasing for x < −1 and 1.75 < x < 3 and increasing for 1 < x < 1.75 and x > 3 . From , f is concave down for 0.1 < x < 2.4 and concave up for x < 0.1 and x > 2.4 . MHR • Calculus and Vectors 12 Solutions

410

 Sketch the graph.



The volume of a cylinder is V = πr2h Use V = 900 to find an equation for h in term of r. 900 = π r 2 h 900 h= 2 πr The surface area of a cylinder is S.A. = 2πr2 + 2πrh. Substitute the equation for h into S.A. to find the equation for the surface area in of r.  900  S.A. = 2π r 2 + 2π r  2   πr   900  = 2πr2 + r  

The metal costs $15.50 per m2. 15.50 15.50 = (100)2 (1)2 = 0.00155

The cost is $0.00155/cm2. The function of cost is: C(r) = 0.00155(SA)  900  = 0.0031 π r 2 + r  


411

Find C ′(r) and set it to zero to find the radius at which cost will be optimized.  900  C ′(r) = 0.0031 2π r − 2   r  0 = 2π r 3 − 900 r=

3

450

π

The cost will be minimized when r =

 450  C 3   π 

3

450

π

cm and h = 2 3

450

π

cm.

  2   900    3 450  = 0.0031 π   +  450     π   3   π    =&0.0031(86.005 + 172.011)



=&0.80

The cost of making the can is $0.80. Chapter 1 to 3 Review


For the field to be rectangular, it must be split up into 4 equal rectangles side by side. Let l represent the length of the whole field and w represent the width. Let P represent the amount of fence needed. P = 2l + 5w 6000 = 2l + 5w 6000 − 5w l= 2

The area of the whole field is: A= l× w  6000 − 5w  =  w 2  = 3000w −

5 2 w 2

Find A′(w) and set it to zero to find the width of the field that will optimize area. A′(w) = 3000 − 5w 0 = 3000 − 5w w = 600

The width of the field is 600 m, which is also the width of each plot of land. MHR • Calculus and Vectors 12 Solutions

412

6000 − 5(600) = 1500 2 The length of the whole field is 1500 m.

Since the field is split up into four congruent plots of land, each side length is 375 m. Therefore, the dimensions of the land are 375 m by 600 m.


413

Chapter 3 Solutions 6nbd

Overview 3z723u

More details 2i4a6q

Related Documents g842

Chapter 3 Solutions 6nbd

Solutions Chapter 3 5u1y5v

Chapter 3 Solutions Manual j1bm

Chapter 3 Solutions 6nbd

Chapter 3 Solutions 6nbd

Chapter 3 Solutions 6nbd

More Documents from "Dan Dinh" 63416x

Bangkok 3mg5h

Chapter 3 Solutions 6nbd

52544_a Nfs2-3030 Installation Manual 4n1w4i

Reglamento Interno De Trabajo 4eq1b

Cateye Cordless 2 Manual 1y442f