Calculus 1 Quiz: Intro To Optimization

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Intro To OptimizationQuestion 1 of 20

The function $f(x) = x^4 - 4x^3 + 10$ has local extrema. What is the absolute minimum value of the function on the interval $[-1, 4]$ ?

-17

-27

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Calculus 1 Quiz

Calculus 1 Quiz: Intro To Optimization

Practice Intro To Optimization in Calculus 1 with focused quiz questions that help you check what you know, review explanations, and build confidence with test-style prompts.

What this quiz covers

This quiz focuses on Intro To Optimization, giving you a quick way to practice the rules, question types, and explanations that matter most for Calculus 1.

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Try each quiz question before looking at the correct answer. Use the explanations to review missed ideas, then come back to similar questions until the pattern feels familiar.

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Question 1

The function $f(x) = x^4 - 4x^3 + 10$ has local extrema. What is the absolute minimum value of the function on the interval $[-1, 4]$ ?

-17 (correct answer)
10
15
-27

Explanation: To find the absolute minimum on a closed interval, we use the Closed Interval Method. First, find the critical points by taking the derivative:

f'(x) = 4x^3 - 12x^2 = 4x^2(x-3)

. Setting

f'(x) = 0

gives critical points at

x=0

and

x=3

. Both are in the interval

[-1, 4]

. Now, we evaluate the function at the critical points and the endpoints:

f(-1) = (-1)^4 - 4(-1)^3 + 10 = 1 - 4(-1) + 10 = 1 + 4 + 10 = 15

f(0) = (0)^4 - 4(0)^3 + 10 = 10

f(3) = (3)^4 - 4(3)^3 + 10 = 81 - 4(27) + 10 = 81 - 108 + 10 = -17

f(4) = (4)^4 - 4(4)^3 + 10 = 256 - 4(64) + 10 = 256 - 256 + 10 = 10

. Comparing the values

{15, 10, -17, 10}

, the absolute minimum is -17. A: Correct. B: This is the value at the critical point

x=0

and the endpoint

x=4

. It is a local maximum or inflection point value, but not the absolute minimum. C: This is the value at the endpoint

x=-1

, which is the absolute maximum on the interval. D: This is the value of

f(3)

with a sign error in calculation:

81-108+10 = -17

. Maybe

81-108 = -27

and the student forgot to add 10.

Question 2

An open-top box is made by cutting congruent squares from the corners of a 12-inch by 12-inch sheet of tin and bending up the sides. What is the maximum volume of such a box?

64 in³
100 in³
128 in³ (correct answer)
144 in³

Explanation: Let

x

be the side length of the square cut from each corner. The dimensions of the resulting box will be: length

L = 12 - 2x

, width

W = 12 - 2x

, and height

H = x

. The volume is

V(x) = LWH = (12 - 2x)(12 - 2x)x = (144 - 48x + 4x^2)x = 4x^3 - 48x^2 + 144x

. The domain for

x

(0, 6)

. To find the maximum volume, we differentiate:

V'(x) = 12x^2 - 96x + 144

. Set

V'(x) = 0

and solve for

x

12(x^2 - 8x + 12) = 0

. This factors as

12(x-2)(x-6) = 0

. The critical points are

x=2

and

x=6

. Since

x=6

is an endpoint of the domain and results in a volume of 0, we only consider

x=2

. Let's check with the second derivative:

V''(x) = 24x - 96

V''(2) = 48 - 96 = -48 < 0

, so

x=2

gives a maximum. The maximum volume is

V(2) = (12 - 2(2))^2(2) = (12 - 4)^2(2) = 8^2(2) = 64(2) = 128

in³. A: This is the area of the base of the optimal box,

(12-4)^2=64

. Students might forget to multiply by the height. B: This is the volume if squares of side length 1 inch are cut out,

V(1)=(10)^2(1)=100

. D: This is the area of the original sheet of tin,

12 \times 12 = 144

Question 3

A rectangle is constructed with its base on the x-axis and its two upper vertices on the parabola $y = 12 - x^2$ . What is the maximum possible perimeter of such a rectangle?

32
26 (correct answer)
24
16

Explanation: Let the upper right vertex of the rectangle be at

(x, y)

, where

y = 12 - x^2

. The width of the rectangle is

2x

and the height is

y = 12 - x^2

. The perimeter

P

is given by

P(x) = 2(2x) + 2(12 - x^2) = 4x + 24 - 2x^2

, for

x

[0, \sqrt{12}]

. To find the maximum perimeter, we take the derivative:

P'(x) = 4 - 4x

. Setting

P'(x) = 0

gives

4 - 4x = 0

, so

x = 1

. To find the maximum value, we check the critical point and the endpoints:

P(1) = 4(1) + 24 - 2(1)^2 = 26

P(0) = 24

P(\sqrt{12}) = 4\sqrt{12} = 8\sqrt{3} \approx 13.86

. The maximum value is 26. A: This is the maximum area of the rectangle, found by maximizing

A(x) = 2x(12-x^2)

, which occurs at

x=2

, giving

A(2)=32

. This is a common error of optimizing the wrong quantity. C: This is the perimeter of the rectangle when its area is maximized (at

x=2

P(2) = 24

. It is also the perimeter at the endpoint

x=0

. D: This is the height (11) plus the width (2) plus a small value, or possibly an arithmetic error. A rectangle of perimeter 16 would have dimensions like

x=2, y=6

, which would be

P=2(4)+2(6)=20

, so this is likely just a calculation error.

Question 4

A particle moves along the x-axis with velocity function $v(t) = t^3 - 6t^2 + 9t + 1$ for $t \ge 0$ . What is the maximum acceleration of the particle on the time interval $t \in [0, 4]$ ?

1
5
9 (correct answer)
12

Explanation: The acceleration function

a(t)

is the derivative of the velocity function

v(t)

. So,

a(t) = v'(t) = 3t^2 - 12t + 9

. To find the maximum acceleration on the closed interval

[0, 4]

, we need to find the critical points of

a(t)

and compare the values of

a(t)

at the critical points and the endpoints. The derivative of acceleration is

a'(t) = 6t - 12

. Setting

a'(t) = 0

gives

6t - 12 = 0

, so

t = 2

. Now we evaluate

a(t)

at the critical point

t=2

and the endpoints

t=0

and

t=4

a(0) = 9

a(2) = 12 - 24 + 9 = -3

a(4) = 48 - 48 + 9 = 9

. The maximum value is 9. A: This is the minimum velocity on the interval,

v(3) = 1

. B: This is the maximum velocity on the interval. A student might optimize velocity instead of acceleration. C: Correct. D: This is the magnitude of the jerk (rate of change of acceleration) at the endpoints:

|a'(0)| = |a'(4)| = 12

Question 5

A cylindrical can with a top and bottom must hold a volume of $1000$ cm $^3$ . The material for the top and bottom costs three times as much per unit area as the material for the side. What is the ratio of the height, $h$ , to the radius, $r$ , that minimizes the cost of the can?

$h/r = 2$
$h/r = 3$
$h/r = 6$ (correct answer)
$h/r = 2/3$

Explanation: Let the cost per unit area for the side be

k

. Then the cost per unit area for the top and bottom is $3k$. The surface area of the side is

2\pi r h

and the area of the top and bottom combined is

2\pi r^2

. The cost function is

C = k(2\pi r h) + 3k(2\pi r^2) = 2\pi k (rh + 3r^2)

. The volume constraint is

V = \pi r^2 h = 1000

, so

h = 1000/(\pi r^2)

. Substituting into the cost function:

C(r) = 2\pi k (r(1000/(\pi r^2)) + 3r^2) = 2\pi k (1000/(\pi r) + 3r^2)

. To find the minimum, we differentiate with respect to

r

C'(r) = 2\pi k (-1000/(\pi r^2) + 6r)

. Setting

C'(r) = 0

gives

6r = 1000/(\pi r^2)

, which simplifies to

6\pi r^3 = 1000

. Substitute

1000 = \pi r^2 h

6\pi r^3 = \pi r^2 h

. Dividing by

\pi r^2

(since

r\neq0

) gives

6r = h

, so the ratio is

h/r = 6

Question 6

A rectangle is formed with vertices at $(0,0)$ , $(x,0)$ , $(x,y)$ , and $(0,y)$ , where the point $(x,y)$ lies on the curve $y = e^{-x/2}$ for $x \ge 0$ . What is the maximum possible area of such a rectangle?

$2/e$ (correct answer)
$1/e$
$1/\sqrt{e}$
$2$

Explanation: The area of the rectangle is given by

A(x) = x \cdot y = x e^{-x/2}

. The domain is

x \ge 0

. To find the maximum area, we compute the derivative of

A(x)

using the product rule:

A'(x) = (1)e^{-x/2} + x(-1/2)e^{-x/2} = e^{-x/2}(1 - x/2)

. Setting

A'(x) = 0

, we find

1 - x/2 = 0

, which gives

x=2

. To confirm this is a maximum, we can use the First Derivative Test. For

0 < x < 2

A'(x) > 0

(area is increasing). For

x > 2

A'(x) < 0

(area is decreasing). Thus, a maximum occurs at

x=2

. The maximum area is

A(2) = 2e^{-2/2} = 2e^{-1} = 2/e

Question 7

A person in a boat is 3 km from the nearest point on a straight shoreline. They wish to reach a house 8 km down the shore from that point. They can row at 6 km/h and run along the shore at 10 km/h. To minimize the travel time, to what point on the shore should they row? Let $x$ be the distance in km from the point on the shore nearest the boat to the point where they land.

$x = 4.00$ km
$x = 3.75$ km
$x = 2.25$ km (correct answer)
$x = 5.00$ km

Explanation: This is a classic optimization problem where you need to minimize travel time by finding the optimal combination of two different speeds. When you see problems involving "shortest time" or "minimum cost" with multiple rates, think calculus optimization. Let's set up the problem systematically. If the person rows to a point

x

km along the shore, they travel

\sqrt{x^2 + 9}

km by boat (using the Pythagorean theorem), then run the remaining

(8-x)

km along shore. The total time function is:

T(x) = \frac{\sqrt{x^2 + 9}}{6} + \frac{8-x}{10}

To minimize time, take the derivative and set it equal to zero:

T'(x) = \frac{x}{6\sqrt{x^2 + 9}} - \frac{1}{10} = 0

Solving:

\frac{x}{6\sqrt{x^2 + 9}} = \frac{1}{10}

Cross-multiply:

10x = 6\sqrt{x^2 + 9}

Square both sides:

100x^2 = 36(x^2 + 9)

This gives us

64x^2 = 324

, so

x^2 = \frac{324}{64} = \frac{81}{16}

Therefore

x = \frac{9}{4} = 2.25

km, which is answer C. Answer A (4.00 km) likely comes from incorrectly setting up the distance relationship. Answer B (3.75 km) might result from algebraic errors when solving the optimization equation. Answer D (5.00 km) could come from misunderstanding the constraint that

x

must be less than 8. Strategy tip: In optimization problems, always verify your derivative algebra carefully—small errors in solving

f'(x) = 0

lead to wrong answers that often appear as distractors.

Question 8

A rectangular poster is to have a total area of 180 square inches. The poster consists of a central printed area with a margin of 1 inch on each side and 1.5 inches on the top and bottom. What are the outer dimensions (width and height) of the poster that will maximize the printed area?

Width = $2\sqrt{30}-2$ in, Height = $3\sqrt{30}-3$ in
Width = $12$ in, Height = $15$ in
Width = $2\sqrt{30}$ in, Height = $3\sqrt{30}$ in (correct answer)
Width = $10$ in, Height = $18$ in

Explanation: This is a classic optimization problem where you need to set up variables, express constraints, and use calculus to find maximum values. When you see "maximize" or "minimize" with given constraints, think about setting up a function to optimize. Let's define the outer dimensions as width

w

and height

h

. Since the total area is 180 square inches, we have

wh = 180

, so

h = \frac{180}{w}

. The printed area is what remains after subtracting the margins. With 1-inch margins on each side, the printed width is

(w-2)

. With 1.5-inch margins on top and bottom, the printed height is

(h-3)

. So the printed area is

A = (w-2)(h-3)

. Substituting our constraint:

A = (w-2)\left(\frac{180}{w}-3\right) = (w-2)\frac{180-3w}{w} = \frac{(w-2)(180-3w)}{w}

To maximize, take the derivative and set it equal to zero. Using the quotient rule:

A'(w) = \frac{w(-6w+186) - (w-2)(180-3w)}{w^2} = \frac{-3w^2 + 12w + 360}{w^2}

Setting

A'(w) = 0

-3w^2 + 12w + 360 = 0

, which gives

w^2 - 4w - 120 = 0

. Using the quadratic formula:

w = \frac{4 \pm \sqrt{16 + 480}}{2} = \frac{4 \pm \sqrt{496}}{2} = 2 \pm \sqrt{124} = 2 \pm 2\sqrt{31}

. Since

w > 0

, we get

w = 2 + 2\sqrt{30} = 2\sqrt{30}

and

h = 3\sqrt{30}

. Choice A gives dimensions that are 2 inches less than the correct values. Choice B uses round numbers that don't satisfy the optimization conditions. Choice D also uses convenient numbers but yields suboptimal printed area. When solving optimization problems, always verify your constraint equation and remember that the derivative test finds critical points—don't forget to check that you're truly maximizing.

Question 9

The profit $P(x)$ from manufacturing $x$ units of a product is modeled by a function for $x>0$ . It is known that $P'(x)$ is continuous, $P'(1000) = 0$ , and the second derivative $P''(x) < 0$ for all $x>0$ . Which of the following must be true?

The profit is maximized at a production level other than $x=1000$ .
The profit is maximized at the production level $x=1000$ . (correct answer)
The profit is minimized at the production level $x=1000$ .
There is not enough information to determine where the profit is maximized.

Explanation: We are given that

x=1000

is a critical point of the profit function because

P'(1000)=0

. We are also given that the second derivative

P''(x)

is negative for all

x>0

. A negative second derivative indicates that the function is concave down. Since

P''(1000)<0

, the Second Derivative Test tells us that

x=1000

corresponds to a local maximum. Because the function is concave down over its entire domain

(0, \infty)

, any local maximum must also be the absolute maximum. Therefore, the profit is maximized at the production level

x=1000

Question 10

An open-topped cylindrical container is to be made from $300\pi$ square inches of material. What is the relationship between the height $h$ and the radius $r$ of the container with the largest possible volume?

$h = \pi r$
$h = 2r$
$h = r/2$
$h = r$ (correct answer)

Explanation: This is a classic optimization problem where you need to maximize volume subject to a constraint on surface area. When you see "largest possible" or "minimum cost" with geometric constraints, you're looking at an optimization problem that requires calculus. To solve this, start by setting up your equations. The surface area constraint is

300\pi = \pi r^2 + 2\pi rh

(bottom plus curved sides), which gives you

h = \frac{300 - r^2}{2r}

. The volume to maximize is

V = \pi r^2 h = \pi r^2 \cdot \frac{300 - r^2}{2r} = \frac{\pi r(300 - r^2)}{2}

. Taking the derivative and setting it equal to zero:

\frac{dV}{dr} = \frac{\pi(300 - 3r^2)}{2} = 0

. This gives

r^2 = 100

, so

r = 10

. Substituting back:

h = \frac{300 - 100}{20} = 10

. Therefore,

h = r

. Looking at the wrong answers: Choice A (

h = \pi r

) would give an irrational relationship that doesn't emerge from this optimization. Choice B (

h = 2r

) is a common trap - it's the relationship for maximizing surface area given fixed volume, not the reverse. Choice C (

h = r/2

) represents the opposite error, perhaps from confusing which variable should be larger. Study tip: In optimization problems, always clearly identify what you're maximizing/minimizing versus what's constrained. Set up both equations before taking derivatives, and remember that geometric optimization often yields surprisingly simple integer or rational relationships.

Question 11

A company's profit $P(x)$ from selling $x$ units of a product is given by $P(x) = R(x) - C(x)$ , where $R(x)$ is revenue and $C(x)$ is cost. The demand function is $p(x) = 500 - 0.5x$ , where $p$ is the price per unit, and the cost function is $C(x) = 10000 + 50x + 0.1x^2$ . Which production level $x$ maximizes profit?

$500$
$375$ (correct answer)
$450$
$250$

Explanation: The revenue function is

R(x) = x \cdot p(x) = x(500 - 0.5x) = 500x - 0.5x^2

. The profit function is

P(x) = R(x) - C(x) = (500x - 0.5x^2) - (10000 + 50x + 0.1x^2) = -0.6x^2 + 450x - 10000

. To find the maximum profit, we differentiate

P(x)

and set it to zero:

P'(x) = -1.2x + 450 = 0

. Solving for

x

gives

1.2x = 450

, so

x = 450/1.2 = 4500/12 = 375

. The second derivative is

P''(x) = -1.2

, which is negative, confirming that

x=375

yields a maximum profit.

Question 12

At $t=0$ , a ship is 50 km north of a lighthouse and is sailing south at 20 km/h. At the same time, another ship is at the lighthouse and sailing east at 15 km/h. At what time $t>0$ are the two ships closest to each other?

$1.6$ hours (correct answer)
$2.0$ hours
$2.5$ hours
$1.2$ hours

Explanation: Let the lighthouse be at the origin

(0,0)

. At time

t

, the first ship's position is

(0, 50-20t)

. The second ship's position is

(15t, 0)

. The square of the distance between them is

S(t) = (15t - 0)^2 + (0 - (50-20t))^2 = 225t^2 + (50-20t)^2 = 225t^2 + 2500 - 2000t + 400t^2 = 625t^2 - 2000t + 2500

. To minimize the distance, we minimize its square. We find the derivative:

S'(t) = 1250t - 2000

. Setting

S'(t) = 0

gives

1250t = 2000

, so

t = 2000/1250 = 200/125 = 8/5 = 1.6

hours. Since

S''(t) = 1250 > 0

, this time corresponds to a minimum distance.

Question 13

The top and bottom margins of a poster are each 2 cm and the side margins are each 1 cm. The area of the printed material on the poster is fixed at 32 cm². What is the minimum possible area of the entire poster?

64 cm²
72 cm² (correct answer)
80 cm²
96 cm²

Explanation: Let the width of the printed material be

w

and the height be

h

. The area of the printed material is

wh = 32

, so

h = 32/w

. The total width of the poster is

W = w + 1 + 1 = w + 2

. The total height of the poster is

H = h + 2 + 2 = h + 4

. The total area of the poster is

A = WH = (w+2)(h+4)

. Substituting for

h

, we get

A(w) = (w+2)(32/w + 4) = 32 + 4w + 64/w + 8 = 4w + 64/w + 40

. To minimize the area, we take the derivative with respect to

w

A'(w) = 4 - 64/w^2

. Setting

A'(w) = 0

gives

4 = 64/w^2

, so

w^2 = 16

, and

w = 4

(since width must be positive). The corresponding height of the printed area is

h = 32/4 = 8

. The minimum poster area is

A(4) = 4(4) + 64/4 + 40 = 16 + 16 + 40 = 72

cm². A: This is twice the printed area, a possible guess. It is also the value of

4w+64/w

w=8

, a potential miscalculation. C: This could result from an error in setting up the dimensions, for example using

W=w+1, H=h+2

and solving. D: This is three times the printed area, another possible guess without using calculus.

Question 14

A rectangular box with a square base and an open top is to have a volume of 32,000 cm³. Find the minimum surface area of the box.

1200 cm²
2400 cm²
4800 cm² (correct answer)
6000 cm²

Explanation: Let the side length of the square base be

x

and the height be

h

. The volume is

V = x^2 h = 32000

, so

h = 32000/x^2

. The surface area of the open-top box consists of the base and four sides:

A = x^2 + 4xh

. Substituting for

h

, we get

A(x) = x^2 + 4x(32000/x^2) = x^2 + 128000/x

. To minimize the area, we take the derivative:

A'(x) = 2x - 128000/x^2

. Setting

A'(x) = 0

gives

2x = 128000/x^2

, so

2x^3 = 128000

, which means

x^3 = 64000

. Taking the cube root, we get

x = 40

cm. The corresponding height is

h = 32000/(40^2) = 32000/1600 = 20

cm. The minimum surface area is

A(40) = 40^2 + 128000/40 = 1600 + 3200 = 4800

cm². A: This could be the area of the four sides

4xh = 4(40)(20) = 3200

, or the base area

x^2=1600

. This is likely an arithmetic error. B: This might result from forgetting the base area and making an error, or from using an incorrect formula for volume or area. C: This is the correct minimum surface area. D: This value may come from an algebraic mistake, such as setting

x=20

and

h=80

, which gives a surface area of

20^2+4(20)(80) = 400+6400=6800

. A value of 6000 is a plausible miscalculation.

Question 15

Let $f$ be a function such that $f'(x) = (x-1)(x-4)^2$ . To find the local extrema of $f$ , a student calculates $f''(x) = (x-4)^2 + 2(x-1)(x-4) = (x-4)(3x-6)$ . The student notes that $f''(1) = 0$ and concludes the Second Derivative Test is inconclusive at $x=1$ . Which of the following is true about $f$ at $x=1$ ?

$f$ has a local maximum at $x=1$ .
$f$ has a local minimum at $x=1$ . (correct answer)
$f$ has neither a local maximum nor a local minimum at $x=1$ .
The test is inconclusive, so no determination can be made.

Explanation: The student correctly found that the Second Derivative Test is inconclusive at

x=1

because

f''(1)=0

. In this case, we must use the First Derivative Test. The critical points are where

f'(x) = (x-1)(x-4)^2 = 0

, which are

x=1

and

x=4

. We need to check the sign of

f'(x)

around

x=1

. The term

(x-4)^2

is always non-negative. For

x < 1

, the term

(x-1)

is negative, so

f'(x)

is negative. For

x

slightly greater than 1 (but less than 4), the term

(x-1)

is positive, so

f'(x)

is positive. Since

f'(x)

changes sign from negative to positive at

x=1

, the function

f

has a local minimum at

x=1

. A: This would be true if the sign of

f'(x)

changed from positive to negative. C: This would be true if the sign of

f'(x)

did not change at

x=1

(e.g., if the factor was

(x-1)^2

). D: While the Second Derivative Test is inconclusive, the First Derivative Test is not. This statement is a common misconception that if one test fails, no conclusion can be drawn.

Question 16

A trough is to be made from a long rectangular piece of sheet metal 30 cm wide by bending up one-third of the metal on each side through an angle $\theta$ . To maximize the cross-sectional area of the trough, what must be the value of $\cos(\theta)$ ?

1/2 (correct answer)
1
$\sqrt{3}/2$
0

Explanation: The sheet metal is 30 cm wide, and is bent up into three sections of 10 cm each. The cross-section is a trapezoid. The base is 10 cm. The two slanted sides are 10 cm. Let

\theta

be the angle with the vertical. The question says angle

\theta

but doesn't specify with what. Let's assume it's the angle the sides make with the horizontal. The height of the trapezoid is

h = 10\sin(\theta)

. The top base is

10 + 2(10\cos(\theta))

. The area is

A(\theta) = \frac{1}{2}(base_1 + base_2)h = \frac{1}{2}(10 + 10 + 20\cos(\theta))(10\sin(\theta)) = (10 + 10\cos(\theta))(10\sin(\theta)) = 100(\sin(\theta) + \sin(\theta)\cos(\theta))

. We want to maximize this for

\theta \in [0, \pi/2]

A'(\theta) = 100(\cos(\theta) + \cos^2(\theta) - \sin^2(\theta))

. Using

\sin^2(\theta)=1-\cos^2(\theta)

, we get

A'(\theta) = 100(\cos(\theta) + 2\cos^2(\theta) - 1)

. Let

u=\cos(\theta)

. We need to solve

2u^2+u-1=0

, which factors as

(2u-1)(u+1)=0

. So

u = 1/2

u = -1

. Since

\theta \in [0, \pi/2]

u=\cos(\theta)

must be non-negative. Thus,

\cos(\theta) = 1/2

. This corresponds to

\theta = \pi/3

or 60 degrees. A: Correct. B:

\cos(\theta)=1

means

\theta=0

, which gives zero area (a flat sheet). C:

\cos(\theta)=\sqrt{3}/2

corresponds to

\theta=30

degrees. This is a plausible guess but not optimal. D:

\cos(\theta)=0

means

\theta=90

degrees, which forms a rectangle with area

10 \times 10 = 100

. The area for

\theta=60

100(\sin(60)+\sin(60)\cos(60)) = 100(\sqrt{3}/2 + (\sqrt{3}/2)(1/2)) = 100(3\sqrt{3}/4) = 75\sqrt{3} \approx 130 > 100

. So the rectangle is not optimal.

Question 17

A function $g(x)$ is continuous on the interval $[0, 10]$ and has a single critical point at $x=4$ . Which of the following pieces of information is sufficient to conclude that $g(x)$ has an absolute minimum at $x=4$ ?

$g''(4) = 5$
$g(4) < g(0)$ and $g(4) < g(10)$ (correct answer)
$g'(x) < 0$ for all $x$ in the interval $(0, 4)$
$g(0) > g(10)$

Explanation: To find the absolute minimum on a closed interval, we must check the function's values at the critical points and the endpoints. The candidates for the location of the absolute minimum are

x=0

x=4

, and

x=10

. B: The statement

g(4) < g(0)

and

g(4) < g(10)

directly compares the value of the function at the only critical point with the values at both endpoints. Since

g(4)

is smaller than the values at the endpoints, and there are no other critical points where a minimum could occur,

g(4)

must be the absolute minimum value. This is a sufficient condition. A: The condition

g''(4) = 5 > 0

is the Second Derivative Test, which only implies that

g(x)

has a local minimum at

x=4

. It is possible that the value of the function at one of the endpoints,

g(0)

g(10)

, is less than

g(4)

. Therefore, this is not sufficient. C: The condition that

g'(x) < 0

for

x

(0, 4)

means the function is decreasing as it approaches

x=4

from the left. This does not tell us anything about the behavior of the function for

x > 4

. The function could continue to decrease, in which case

x=4

would not be a minimum. This is not sufficient. D: Comparing the values at the endpoints does not provide enough information about the value at the critical point

x=4

. This is not sufficient.

Question 18

The illumination from a light source is directly proportional to the strength of the source and inversely proportional to the square of the distance from the source. Two light sources, one of which is 8 times as strong as the other, are placed 12 feet apart. At what distance from the stronger light source is the total illumination at a minimum?

4 feet
6 feet
8 feet (correct answer)
9 feet

Explanation: Let the strength of the weaker source be

S

and the stronger source be

8S

. Let a point be at a distance

x

from the stronger source. Then its distance from the weaker source is

12-x

. The total illumination

I

at this point is the sum of the illuminations from both sources:

I(x) = \frac{k(8S)}{x^2} + \frac{kS}{(12-x)^2}

, where

k

is a proportionality constant. We can minimize

f(x) = I(x)/(kS) = \frac{8}{x^2} + \frac{1}{(12-x)^2}

for

x \in (0, 12)

. We differentiate:

f'(x) = -\frac{16}{x^3} - \frac{-2}{(12-x)^3} = -\frac{16}{x^3} + \frac{2}{(12-x)^3}

. Setting

f'(x) = 0

gives

\frac{16}{x^3} = \frac{2}{(12-x)^3}

. This simplifies to

8(12-x)^3 = x^3

. Taking the cube root of both sides gives

2(12-x) = x

. So,

24 - 2x = x

, which gives

3x = 24

, and

x = 8

. The minimum illumination occurs at 8 feet from the stronger light source. A: This would mean

x=4

and

12-x=8

. This results if the cube root is not taken correctly, leading to

2(12-x)=x

24-2x=x

x=8

. Wait. This works. What if they solve

8(12-x)=x

96-8x=x

9x=96

. No. What if they take square root instead of cube root?

\sqrt{8}(12-x)=x

2\sqrt{2}(12-x)=x

. This is too messy. A distance of 4 feet implies

x=4

. Then

12-x=8

. Then

16/4^3 = 16/64 = 1/4

and

2/8^3 = 2/512 = 1/256

. Not equal. So 4 feet is incorrect. B: 6 feet is halfway, which would be the minimum if the sources had equal strength. D: 9 feet would put the point 3 feet from the weaker source. This might be a ratio error.

Question 19

Let $f(x) = x^3 + ax^2 + bx + c$ . The function $f$ has a local maximum at $x=-2$ and a local minimum at $x=4$ . What is the value of $b$ ?

-24 (correct answer)
-12
3
8

Explanation: For a function to have local extrema at given points, its derivative must be zero at those points. The derivative of

f(x)

f'(x) = 3x^2 + 2ax + b

. We are given that

x=-2

and

x=4

are critical points, so they must be the roots of

f'(x) = 0

. Therefore, we can write

f'(x)

in factored form:

f'(x) = k(x - (-2))(x - 4) = k(x+2)(x-4)

for some constant

k

. Expanding this gives

f'(x) = k(x^2 - 2x - 8) = kx^2 - 2kx - 8k

. By comparing the coefficients of this form with

f'(x) = 3x^2 + 2ax + b

, we can see that the coefficient of

x^2

must match, so

k=3

. Now we can find the other coefficients:

2a = -2k = -2(3) = -6

, so

a=-3

. And

b = -8k = -8(3) = -24

. The value of

b

is -24. A: Correct. B: This may result from a sign error or using

k=3/2

. C: This is the value of

-a

. D: This could be from

-2 \times -4

, ignoring the constant

k

Question 20

A cylindrical can with a closed top and bottom is to be constructed to hold a volume of $16\pi$ cubic centimeters. The material for the top and bottom costs $2 per square centimeter, and the material for the side costs $1 per square centimeter. What is the ratio of height to radius, $h/r$ , for the can with the minimum cost?

1
2
4 (correct answer)
8

Explanation: The volume is

V = \pi r^2 h = 16\pi

, so

h = 16/r^2

. The cost function

C

is given by the cost of the top and bottom plus the cost of the side:

C = 2(2\pi r^2) + 1(2\pi rh) = 4\pi r^2 + 2\pi rh

. Substituting for

h

, we get

C(r) = 4\pi r^2 + 2\pi r(16/r^2) = 4\pi r^2 + 32\pi/r

. To minimize cost, we find the derivative:

C'(r) = 8\pi r - 32\pi/r^2

. Setting

C'(r) = 0

gives

8\pi r = 32\pi/r^2

, which simplifies to

r^3 = 4

, so

r = \sqrt[3]{4}

. The corresponding height is

h = 16/r^2 = 16/(\sqrt[3]{4})^2 = 16/4^{2/3} = 4^2/4^{2/3} = 4^{4/3}

. The ratio is

h/r = 4^{4/3} / 4^{1/3} = 4^{(4/3 - 1/3)} = 4^1 = 4

. A: This ratio would be found if the costs were equal and the quantity to be optimized was

r+h

or something similar, likely an algebraic error. B: This is the well-known ratio that minimizes the surface area of a cylinder for a fixed volume, a common trap for students who have memorized that result without considering the differential costs. D: This ratio may result from an algebraic error, such as setting

r^3=8

instead of

r^3=4