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Math  /  Algebra

QuestionExercise 16B 1 The profit, $P\$ P million, made by a business which invests $x\$ x million in advertising is given by P=10x2−10x4 for x⩾0P=10 x^{2}-10 x^{4} \text { for } x \geqslant 0
Find the maximum profit the company can make based on this model. 2 A manufacturer produces smartphone covers. They know that if they sell nn thousand covers, they will make a profit of $P\$ P hundred, and they use the model P=20n−3n2−n5P=20 n-3 n^{2}-n^{5}. Find, to the nearest dollar, the maximum profit they can make according to this model. 3 The fuel consumption of a car, FF litres per 100 km , varies with the speed, v km h−1v \mathrm{~km} \mathrm{~h}^{-1}, according to the equation F=(3×10−6)v3−(1.2×10−4)v2−0.035v+12F=\left(3 \times 10^{-6}\right) v^{3}-\left(1.2 \times 10^{-4}\right) v^{2}-0.035 v+12 At what speed should the car be driven in order to minimize fuel consumption? 4 The rate of growth, RR, of a population of bacteria, tt hours after the start of an experiment, is modelled by R=6t−47t4R=\frac{6}{t}-\frac{47}{t^{4}} for t⩾2t \geqslant 2. Find the time when the population growth is the fastest. 5 A rectangle has width x cmx \mathrm{~cm} and length 20−x cm20-x \mathrm{~cm}. a Find the perimeter of the rectangle. b Find the maximum possible area of the rectangle. 6 A rectangle has sides 3x cm3 x \mathrm{~cm} and 14x cm\frac{14}{x} \mathrm{~cm}. a Find the area of the rectangle. b Find the smallest possible perimeter of the rectangle. 7 A cuboid is formed by a square base of side length x cmx \mathrm{~cm}. The other side of the cuboid is of length 9−x cm9-x \mathrm{~cm}. Find the maximum possible volume of the cuboid. 8 A rectangle has area 36 cm236 \mathrm{~cm}^{2}. Let x cmx \mathrm{~cm} be the length of one of the sides. a Express the perimeter of the rectangle in terms of xx. b Hence find the smallest possible perimeter.

Studdy Solution

STEP 1

1. We are given multiple problems involving optimization of functions.
2. Each problem requires finding either a maximum or minimum value.
3. Calculus, specifically differentiation, will be used to find critical points.
4. The critical points will be evaluated to determine if they correspond to maxima or minima.

STEP 2

1. Problem 1: Maximize the profit function P=10x2−10x4 P = 10x^2 - 10x^4 .
2. Problem 2: Maximize the profit function P=20n−3n2−n5 P = 20n - 3n^2 - n^5 .
3. Problem 3: Minimize the fuel consumption function F=(3×10−6)v3−(1.2×10−4)v2−0.035v+12 F = (3 \times 10^{-6})v^3 - (1.2 \times 10^{-4})v^2 - 0.035v + 12 .
4. Problem 4: Maximize the growth rate function R=6t−47t4 R = \frac{6}{t} - \frac{47}{t^4} .
5. Problem 5: a) Find the perimeter of a rectangle. b) Maximize the area of the rectangle.
6. Problem 6: a) Find the area of a rectangle. b) Minimize the perimeter of the rectangle.
7. Problem 7: Maximize the volume of a cuboid.
8. Problem 8: a) Express the perimeter of a rectangle in terms of x x . b) Minimize the perimeter.

STEP 3

Differentiate the profit function P=10x2−10x4 P = 10x^2 - 10x^4 with respect to x x :
dPdx=20x−40x3 \frac{dP}{dx} = 20x - 40x^3
Set the derivative equal to zero to find critical points:
20x−40x3=0 20x - 40x^3 = 0 20x(1−2x2)=0 20x(1 - 2x^2) = 0
Solve for x x :
x=0or1−2x2=0 x = 0 \quad \text{or} \quad 1 - 2x^2 = 0 2x2=1 2x^2 = 1 x2=12 x^2 = \frac{1}{2} x=±12 x = \pm\frac{1}{\sqrt{2}}
Since x≥0 x \geq 0 , we consider x=12 x = \frac{1}{\sqrt{2}} .

STEP 4

Evaluate the second derivative to determine the nature of the critical points:
d2Pdx2=20−120x2 \frac{d^2P}{dx^2} = 20 - 120x^2
Evaluate at x=12 x = \frac{1}{\sqrt{2}} :
d2Pdx2=20−120(12)2 \frac{d^2P}{dx^2} = 20 - 120\left(\frac{1}{\sqrt{2}}\right)^2 d2Pdx2=20−120(12) \frac{d^2P}{dx^2} = 20 - 120\left(\frac{1}{2}\right) d2Pdx2=20−60 \frac{d^2P}{dx^2} = 20 - 60 d2Pdx2=−40 \frac{d^2P}{dx^2} = -40
Since the second derivative is negative, x=12 x = \frac{1}{\sqrt{2}} is a maximum.

STEP 5

Calculate the maximum profit:
P=10(12)2−10(12)4 P = 10\left(\frac{1}{\sqrt{2}}\right)^2 - 10\left(\frac{1}{\sqrt{2}}\right)^4 P=10(12)−10(14) P = 10\left(\frac{1}{2}\right) - 10\left(\frac{1}{4}\right) P=5−2.5 P = 5 - 2.5 P=2.5 P = 2.5
The maximum profit is $2.5 \$2.5 million.

STEP 6

Differentiate the profit function P=20n−3n2−n5 P = 20n - 3n^2 - n^5 with respect to n n :
dPdn=20−6n−5n4 \frac{dP}{dn} = 20 - 6n - 5n^4
Set the derivative equal to zero to find critical points:
20−6n−5n4=0 20 - 6n - 5n^4 = 0
This equation may require numerical methods or graphing to solve for n n .

STEP 7

Use numerical methods or graphing to find the critical points of n n .
Evaluate the second derivative to determine the nature of the critical points:
d2Pdn2=−6−20n3 \frac{d^2P}{dn^2} = -6 - 20n^3
Evaluate at the critical points to determine maxima or minima.

STEP 8

Calculate the maximum profit using the critical points found.

STEP 9

Differentiate the fuel consumption function F=(3×10−6)v3−(1.2×10−4)v2−0.035v+12 F = (3 \times 10^{-6})v^3 - (1.2 \times 10^{-4})v^2 - 0.035v + 12 with respect to v v :
dFdv=3×10−6⋅3v2−2×1.2×10−4⋅v−0.035 \frac{dF}{dv} = 3 \times 10^{-6} \cdot 3v^2 - 2 \times 1.2 \times 10^{-4} \cdot v - 0.035
Set the derivative equal to zero to find critical points.

STEP 10

Solve the equation for v v using numerical methods or graphing.
Evaluate the second derivative to determine the nature of the critical points.

STEP 11

Calculate the speed that minimizes fuel consumption using the critical points found.

STEP 12

Differentiate the growth rate function R=6t−47t4 R = \frac{6}{t} - \frac{47}{t^4} with respect to t t :
dRdt=−6t2+188t5 \frac{dR}{dt} = -\frac{6}{t^2} + \frac{188}{t^5}
Set the derivative equal to zero to find critical points.

STEP 13

Solve the equation for t t using algebraic manipulation.
Evaluate the second derivative to determine the nature of the critical points.

STEP 14

Calculate the time when the population growth is the fastest using the critical points found.

STEP 15

a) Calculate the perimeter of the rectangle with width x x cm and length 20−x 20-x cm:
Perimeter=2(x+(20−x))=40 cm \text{Perimeter} = 2(x + (20-x)) = 40 \text{ cm}
b) Calculate the area of the rectangle:
Area=x(20−x)=20x−x2 \text{Area} = x(20-x) = 20x - x^2
Differentiate the area function with respect to x x :
d(Area)dx=20−2x \frac{d(\text{Area})}{dx} = 20 - 2x
Set the derivative equal to zero to find critical points:
20−2x=0 20 - 2x = 0 x=10 x = 10

STEP 16

Evaluate the second derivative to determine the nature of the critical points:
d2(Area)dx2=−2 \frac{d^2(\text{Area})}{dx^2} = -2
Since the second derivative is negative, x=10 x = 10 is a maximum.
Calculate the maximum area:
Area=10(20−10)=100 cm2 \text{Area} = 10(20-10) = 100 \text{ cm}^2

STEP 17

a) Calculate the area of the rectangle with sides 3x 3x cm and 14x \frac{14}{x} cm:
Area=3x⋅14x=42 cm2 \text{Area} = 3x \cdot \frac{14}{x} = 42 \text{ cm}^2
b) Calculate the perimeter of the rectangle:
Perimeter=2(3x+14x) \text{Perimeter} = 2(3x + \frac{14}{x})
Differentiate the perimeter function with respect to x x :
d(Perimeter)dx=6−28x2 \frac{d(\text{Perimeter})}{dx} = 6 - \frac{28}{x^2}
Set the derivative equal to zero to find critical points:
6−28x2=0 6 - \frac{28}{x^2} = 0 28x2=6 \frac{28}{x^2} = 6 x2=286 x^2 = \frac{28}{6} x2=143 x^2 = \frac{14}{3} x=143 x = \sqrt{\frac{14}{3}}

STEP 18

Evaluate the second derivative to determine the nature of the critical points:
d2(Perimeter)dx2=56x3 \frac{d^2(\text{Perimeter})}{dx^2} = \frac{56}{x^3}
Since the second derivative is positive, x=143 x = \sqrt{\frac{14}{3}} is a minimum.
Calculate the minimum perimeter:
Perimeter=2(3143+14143) \text{Perimeter} = 2\left(3\sqrt{\frac{14}{3}} + \frac{14}{\sqrt{\frac{14}{3}}}\right)
Simplify and calculate the value.

STEP 19

Calculate the volume of the cuboid with square base x x cm and height 9−x 9-x cm:
Volume=x2(9−x)=9x2−x3 \text{Volume} = x^2(9-x) = 9x^2 - x^3
Differentiate the volume function with respect to x x :
d(Volume)dx=18x−3x2 \frac{d(\text{Volume})}{dx} = 18x - 3x^2
Set the derivative equal to zero to find critical points:
18x−3x2=0 18x - 3x^2 = 0 3x(6−x)=0 3x(6-x) = 0
Solve for x x :
x=0orx=6 x = 0 \quad \text{or} \quad x = 6

STEP 20

Evaluate the second derivative to determine the nature of the critical points:
d2(Volume)dx2=18−6x \frac{d^2(\text{Volume})}{dx^2} = 18 - 6x
Evaluate at x=6 x = 6 :
d2(Volume)dx2=18−36 \frac{d^2(\text{Volume})}{dx^2} = 18 - 36 d2(Volume)dx2=−18 \frac{d^2(\text{Volume})}{dx^2} = -18
Since the second derivative is negative, x=6 x = 6 is a maximum.
Calculate the maximum volume:
Volume=62(9−6)=108 cm3 \text{Volume} = 6^2(9-6) = 108 \text{ cm}^3

STEP 21

a) Express the perimeter of the rectangle with area 36 cm2 36 \text{ cm}^2 and side length x x cm:
Area=x⋅36x=36 \text{Area} = x \cdot \frac{36}{x} = 36 Perimeter=2(x+36x) \text{Perimeter} = 2(x + \frac{36}{x})
b) Differentiate the perimeter function with respect to x x :
d(Perimeter)dx=2(1−36x2) \frac{d(\text{Perimeter})}{dx} = 2(1 - \frac{36}{x^2})
Set the derivative equal to zero to find critical points:
1−36x2=0 1 - \frac{36}{x^2} = 0 x2=36 x^2 = 36 x=6 x = 6

STEP 22

Evaluate the second derivative to determine the nature of the critical points:
d2(Perimeter)dx2=72x3 \frac{d^2(\text{Perimeter})}{dx^2} = \frac{72}{x^3}
Since the second derivative is positive, x=6 x = 6 is a minimum.
Calculate the minimum perimeter:
Perimeter=2(6+366)=24 cm \text{Perimeter} = 2(6 + \frac{36}{6}) = 24 \text{ cm}
The solutions to the problems are as follows:
1. Maximum profit: $2.5 \$2.5 million.
2. Maximum profit: Solve numerically.
3. Speed for minimum fuel consumption: Solve numerically.
4. Time for fastest growth: Solve algebraically.
5. a) Perimeter: 40 cm 40 \text{ cm} , b) Maximum area: 100 cm2 100 \text{ cm}^2 .
6. a) Area: 42 cm2 42 \text{ cm}^2 , b) Minimum perimeter: Solve numerically.
7. Maximum volume: 108 cm3 108 \text{ cm}^3 .
8. a) Perimeter expression: 2(x+36x) 2(x + \frac{36}{x}) , b) Minimum perimeter: 24 cm 24 \text{ cm} .

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