Calculus 1 Quiz: Solving Optimization Problems

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Solving Optimization ProblemsQuestion 1 of 11

Find the absolute maximum value of the function $f(x) = x^3 - 6x^2 + 9x + 5$ on the interval $[0, 5]$ .

1

9

5

25

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Calculus 1 Quiz

Calculus 1 Quiz: Solving Optimization Problems

Practice Solving Optimization Problems 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 Solving Optimization Problems, giving you a quick way to practice the rules, question types, and explanations that matter most for Calculus 1.

How to use this quiz

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.

All questions

Question 1

Find the absolute maximum value of the function $f(x) = x^3 - 6x^2 + 9x + 5$ on the interval $[0, 5]$ .

$1$
$9$
$5$
$25$ (correct answer)

Explanation: When you need to find the absolute maximum of a continuous function on a closed interval, you must check both critical points inside the interval and the endpoints. This ensures you don't miss the global maximum. First, find the critical points by taking the derivative and setting it equal to zero:

f'(x) = 3x^2 - 12x + 9 = 3(x^2 - 4x + 3) = 3(x-1)(x-3)

Setting

f'(x) = 0

gives critical points at

x = 1

and

x = 3

, both of which lie within

[0,5]

. Now evaluate

f(x)

at the critical points and endpoints:

$f(0) = 0 - 0 + 0 + 5 = 5$
$f(1) = 1 - 6 + 9 + 5 = 9$
$f(3) = 27 - 54 + 27 + 5 = 5$
$f(5) = 125 - 150 + 45 + 5 = 25$

The absolute maximum value is 25, making D correct. Choice A (1) doesn't correspond to any function value at critical points or endpoints. Choice B (9) is the value at the local maximum

x = 1

, but you must check endpoints too. Choice C (5) represents the function values at both endpoints

x = 0

and

x = 3

, but these aren't the absolute maximum. Strategy tip: Always evaluate the function at every critical point within the interval AND both endpoints. The absolute extrema on closed intervals must occur at one of these locations, so checking all candidates systematically prevents missing the answer.

Question 2

A company is designing a cylindrical container with a volume of $16\pi$ cubic meters. The material for the top and bottom lids costs $10 per square meter, and the material for the side costs $2 per square meter. What is the radius, in meters, of the cylinder that minimizes the total cost of the material?

$2/\sqrt[3]{5}$ (correct answer)
$2$
$\sqrt[3]{40}$
$\sqrt[3]{5}/2$

Explanation: Let

r

be the radius and

h

be the height of the cylinder. The volume is

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

, which implies

h = 16/r^2

. The cost function is

C(r,h) = 10(\text{Area of top/bottom}) + 2(\text{Area of side}) = 10(2\pi r^2) + 2(2\pi rh) = 20\pi r^2 + 4\pi rh

. Substituting for

h

, we get the cost as a function of

r

C(r) = 20\pi r^2 + 4\pi r(16/r^2) = 20\pi r^2 + 64\pi/r

. To find the minimum cost, we take the derivative with respect to

r

and set it to zero:

C'(r) = 40\pi r - 64\pi/r^2 = 0

. This gives

40\pi r = 64\pi/r^2

, which simplifies to

40r^3 = 64

, or

r^3 = 64/40 = 8/5

. Thus,

r = \sqrt[3]{8/5} = 2/\sqrt[3]{5}

. The second derivative,

C''(r) = 40\pi + 128\pi/r^3

, is positive for

r>0

, confirming this is a minimum.

Question 3

A continuous function $f(x)$ is defined on the closed interval $[a, b]$ . The function is twice differentiable on $(a,b)$ . If $f'(c) = 0$ for some $c \in (a,b)$ , and $f''(x) > 0$ for all $x \in (a,b)$ , which of the following statements must be true?

The absolute minimum of $f(x)$ occurs at $x=c$ and the absolute maximum occurs at $x=a$ .
The absolute minimum of $f(x)$ occurs at $x=c$ and the absolute maximum occurs at either $x=a$ or $x=b$ . (correct answer)
The absolute minimum of $f(x)$ occurs at $x=a$ and the absolute maximum occurs at $x=c$ .
The absolute minimum of $f(x)$ occurs at either $x=a$ or $x=b$ and the absolute maximum occurs at $x=c$ .

Explanation: The condition

f'(c) = 0

identifies

c

as a critical point. The condition

f''(x) > 0

for all

x \in (a,b)

means the function is concave up on the entire interval. By the Second Derivative Test, since

f''(c) > 0

, the function has a local minimum at

x=c

. Because the function is concave up across the entire interval, this local minimum is also the absolute minimum on

[a,b]

. For a function that is concave up on a closed interval, the absolute maximum must occur at one of the endpoints,

x=a

x=b

. Therefore, the absolute minimum is at

x=c

and the absolute maximum is at either

x=a

x=b

Question 4

A wire of length $L$ is cut into two pieces. One piece is bent into a square, and the other is bent into a circle. What is the ratio of the side length of the square to the radius of the circle that results in the minimum possible total area?

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

Explanation: This is a classic optimization problem where you need to minimize the total area of two shapes formed from a wire of fixed length. When you encounter problems involving "minimum" or "maximum" with constraints, think calculus optimization: set up the function, take the derivative, and find critical points. Let's say the square uses length

x

of wire and the circle uses length

L-x

. The square's side length is

\frac{x}{4}

, so its area is

\frac{x^2}{16}

. The circle's circumference is

L-x = 2\pi r

, so the radius is

r = \frac{L-x}{2\pi}

and its area is

\pi r^2 = \frac{(L-x)^2}{4\pi}

. The total area function is

A(x) = \frac{x^2}{16} + \frac{(L-x)^2}{4\pi}

. Taking the derivative:

A'(x) = \frac{x}{8} - \frac{L-x}{2\pi}

. Setting equal to zero and solving:

\frac{x}{8} = \frac{L-x}{2\pi}

, which gives us

x = \frac{4L}{4+\pi}

. This means the side length is

\frac{L}{4+\pi}

and the radius is

\frac{L}{2\pi+8}

. The ratio of side length to radius is

\frac{L/(4+\pi)}{L/(2\pi+8)} = \frac{2\pi+8}{4+\pi} = 2

. Choice (A)

1/2

would come from incorrectly setting up the ratio. Choice (B)

\pi/2

might result from confusing circumference and radius relationships. Choice (C)

\sqrt{\pi}

could arise from incorrectly handling the area formulas. For optimization problems, always verify your derivative work and double-check that you're finding the ratio requested in the question.

Question 5

A person is on an island 3 miles from the nearest point on a straight shoreline. They wish to reach a house located 8 miles down the shore from that point. The person can row a boat at a rate of 2 mph and can walk at a rate of 4 mph. At what distance from the house should they land the boat to minimize their travel time?

$\sqrt{3}$ miles
$8 - \sqrt{3}$ miles (correct answer)
$0$ miles
$8$ miles

Explanation: This is a classic optimization problem that tests your ability to set up and minimize a function representing total travel time. When you see problems involving two different speeds and finding an optimal point, think about expressing the total time as a function of your decision variable. Let's say you land the boat at a point

x

miles from the house. You'll row from the island to this landing point, then walk the remaining

x

miles to the house. The rowing distance is

\sqrt{3^2 + (8-x)^2} = \sqrt{9 + (8-x)^2}

miles (using the Pythagorean theorem). Your total time function is:

T(x) = \frac{\sqrt{9 + (8-x)^2}}{2} + \frac{x}{4}

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

T'(x) = \frac{(8-x)(-1)}{2\sqrt{9 + (8-x)^2}} + \frac{1}{4} = 0

Solving this equation:

\frac{8-x}{2\sqrt{9 + (8-x)^2}} = \frac{1}{4}

This leads to

2(8-x) = \sqrt{9 + (8-x)^2}

. Squaring both sides and solving gives us

x = 8 - \sqrt{3}

. Choice A (

\sqrt{3}

miles) represents a common algebraic error when solving the optimization equation. Choice C (0 miles) would mean going directly to the nearest shore point, ignoring the walking advantage. Choice D (8 miles) suggests rowing directly to the house, which wastes the faster walking speed. For optimization problems, always set up your constraint equation carefully using geometry, then use calculus to find the critical points. Double-check that your answer makes physical sense.

Question 6

The marginal revenue for a product is fixed at $100 per unit. The marginal cost is $MC(q) = q^2 - 17q + 160 $, where$ q$ is the quantity produced. There are two production levels where marginal revenue equals marginal cost. Which level maximizes the profit?

$8.5$
$5$
$12$ (correct answer)
$17$

Explanation: When you encounter profit optimization problems, remember that profit is maximized where marginal revenue equals marginal cost, but you must also check the second derivative to confirm whether each point is a maximum or minimum. To find where marginal revenue equals marginal cost, set

100 = q^2 - 17q + 160

. Rearranging gives

q^2 - 17q + 60 = 0

. Using the quadratic formula or factoring:

(q - 5)(q - 12) = 0

, so

q = 5

q = 12

. Since we have two critical points, we need the second derivative test to determine which maximizes profit. The profit function's second derivative equals the negative of the marginal cost's first derivative. Taking the derivative of

MC(q) = q^2 - 17q + 160

gives

MC'(q) = 2q - 17

. Therefore, the second derivative of profit is

-(2q - 17) = 17 - 2q

. At

q = 5

17 - 2(5) = 7 > 0

, indicating a local maximum. At

q = 12

17 - 2(12) = -7 < 0

, indicating a local minimum. The answer is (C) 12 because this is where profit is maximized. (A) 8.5 is neither critical point. (B) 5 is where marginal revenue equals marginal cost, but the second derivative test shows this is actually a profit minimum, not maximum. (D) 17 comes from mistaking the coefficient in the marginal cost function for a production level. Study tip: Always use the second derivative test when you find multiple points where marginal revenue equals marginal cost—equal derivatives don't guarantee maximum profit.

Question 7

A rectangle is inscribed under the arch of the parabola $y = 12 - x^2$ with its base on the x-axis. What is the maximum possible perimeter of such a rectangle?

$26$ (correct answer)
$24$
$32$
$24.5$

Explanation: Let the top-right vertex of the rectangle be

(x, y)

, where

y = 12 - x^2

. By symmetry, the width of the rectangle is

2x

and the height is

y

. The perimeter is

P = 2(\text{width}) + 2(\text{height}) = 2(2x) + 2(y) = 4x + 2y

. Substituting for

y

, we get

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

. To maximize the perimeter, we find the derivative and set it to zero:

P'(x) = -4x + 4 = 0

, which gives

x=1

. The second derivative

P''(x) = -4 < 0

, confirming a maximum. The maximum perimeter is

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

Question 8

What is the minimum distance from the point $(4, 0)$ to a point on the curve $y = \sqrt{2x}$ ?

$\sqrt{7}$ (correct answer)
$3$
$7$
$\sqrt{6}$

Explanation: Let

(x, y)

be a point on the curve

y = \sqrt{2x}

. The distance

D

from

(4, 0)

(x, y)

is given by

D = \sqrt{(x-4)^2 + (y-0)^2}

. To simplify calculations, we can minimize the square of the distance,

S = D^2 = (x-4)^2 + y^2

. Since

y = \sqrt{2x}

, we have

y^2 = 2x

. Substituting this into the equation for

S

gives

S(x) = (x-4)^2 + 2x = x^2 - 8x + 16 + 2x = x^2 - 6x + 16

. To find the minimum, we take the derivative and set it to zero:

S'(x) = 2x - 6 = 0

, which yields

x=3

. The second derivative

S''(x) = 2 > 0

, confirming a minimum. The question asks for the minimum distance

D

, not

S

x

. We evaluate

S(3) = 3^2 - 6(3) + 16 = 9 - 18 + 16 = 7

. The minimum distance is

D = \sqrt{S(3)} = \sqrt{7}

Question 9

The demand function for a product is given by $p(q) = \sqrt{4500 - q}$ , where $p$ is the price per unit and $q$ is the number of units sold. What is the quantity $q$ that maximizes the revenue?

$1500$
$2250$
$4500$
$3000$ (correct answer)

Explanation: When you encounter a revenue maximization problem, you need to set up the revenue function and find where its derivative equals zero. Revenue is always price times quantity:

R(q) = p \cdot q

. Given the demand function

p(q) = \sqrt{4500 - q}

, the revenue function becomes:

R(q) = q \cdot \sqrt{4500 - q} = q(4500 - q)^{1/2}

To find the maximum, take the derivative using the product rule:

R'(q) = (4500 - q)^{1/2} + q \cdot \frac{1}{2}(4500 - q)^{-1/2} \cdot (-1)

Simplifying:

R'(q) = \sqrt{4500 - q} - \frac{q}{2\sqrt{4500 - q}}

Setting

R'(q) = 0

and multiplying through by

2\sqrt{4500 - q}

2(4500 - q) - q = 0

9000 - 2q - q = 0

9000 = 3q

q = 3000

Choice A (1500) represents

q = \frac{4500}{3}

, which might come from incorrectly dividing the domain by 3. Choice B (2250) equals

\frac{4500}{2}

, a common error of simply taking half the maximum possible quantity. Choice C (4500) is the upper bound of the domain where

p = 0

, meaning no revenue despite maximum quantity. The key insight is that maximum revenue doesn't occur at the endpoints but at an interior point where marginal revenue equals zero. Always remember: set up

R(q) = p(q) \cdot q

, differentiate, and solve

R'(q) = 0

Question 10

A movie screen on a wall is 10 feet high and its bottom is 4 feet above an observer's eye level. At what distance $x$ from the wall should the observer stand to maximize the viewing angle $\theta$ ?

$\sqrt{18}$ feet
$9$ feet
$\sqrt{56}$ feet (correct answer)
$56$ feet

Explanation: This is a classic optimization problem where you need to find the maximum viewing angle. When you see problems asking to maximize or minimize angles, areas, or distances, think calculus optimization: set up an equation for what you want to optimize, then use derivatives to find critical points. Start by setting up the geometry. Place the observer at distance

x

from the wall. The screen extends from 4 feet to 14 feet above eye level. Using trigonometry, the viewing angle

\theta

satisfies:

\tan(\theta) = \frac{10x}{x^2 + 56}

This comes from the difference of two arctangent functions representing the angles to the top and bottom of the screen. To maximize

\theta

, differentiate and set equal to zero:

\frac{d}{dx}\left(\frac{10x}{x^2 + 56}\right) = \frac{10(56 - x^2)}{(x^2 + 56)^2} = 0

This gives us

56 - x^2 = 0

, so

x^2 = 56

, and

x = \sqrt{56}

. Looking at the wrong answers: A)

\sqrt{18}

likely comes from incorrectly using just the screen height (

\sqrt{4 \times 10} = \sqrt{40}

is close). B) 9 feet might result from arithmetic errors in the optimization process. D) 56 feet is the value of

x^2

, not

x

itself - a common mistake when students forget to take the square root. Study tip: In optimization problems involving angles, always check whether you need the square root of your intermediate result. The derivative will often give you

x^2

, requiring one more step to find

x

Question 11

A rectangular box with a square base and an open top is to be constructed from a sheet of cardboard with a surface area of 192 square inches. What is the maximum possible volume of such a box?

$8$ cubic inches
$128\sqrt{2}$ cubic inches
$256$ cubic inches (correct answer)
$128$ cubic inches

Explanation: This is a classic optimization problem that tests your ability to set up constraints and use calculus to find maximum values. When you see "maximum volume" with material constraints, you'll need to express volume as a function of one variable, then use derivatives to optimize. Let's define the problem systematically. If the square base has side length

x

and the box has height

h

, then the surface area constraint gives us:

x^2 + 4xh = 192

(base plus four sides, no top). Solving for height:

h = \frac{192 - x^2}{4x}

. The volume function becomes:

V(x) = x^2h = x^2 \cdot \frac{192 - x^2}{4x} = \frac{x(192 - x^2)}{4} = 48x - \frac{x^3}{4}

. To find the maximum, take the derivative and set it equal to zero:

V'(x) = 48 - \frac{3x^2}{4} = 0

. Solving:

\frac{3x^2}{4} = 48

, so

x^2 = 64

and

x = 8

. When

x = 8

, we get

h = \frac{192 - 64}{32} = 4

, giving us

V = 8^2 \cdot 4 = 256

cubic inches. Answer A (

8

) likely comes from finding just the side length rather than the volume. Answer B (

128\sqrt{2}

) might result from algebraic errors in the optimization. Answer D (

128

) could come from using incorrect dimensions or miscalculating the final volume. Remember: in optimization problems, always verify your critical point gives a maximum by checking the second derivative or testing nearby values.