AP Calculus BC Practice Quiz: Introduction to Optimization Problems
Written by AP Content Team, Verified for 2026 AP Exams, Last updated: May 2026
Test your understanding with short quizzes. This quiz has 7 questions to check your progress.
Question 1 of 7
All Questions (7)
A) The integral of the function, ∫f(x)dx
B) The derivative of the function, f'(x)
C) The value of the function at the endpoints of the interval
D) The second derivative of the function, f''(x)
Correct Answer: B
According to the provided content, the derivative is used to solve optimization problems. Setting the derivative of a function to zero (f'(x) = 0) or finding where it is undefined allows us to find the critical points, which are the potential locations for local minima and maxima.
A) -1
B) 0
C) 4
D) 5
Correct Answer: C
To find the absolute minimum, we must test the critical points and the endpoints of the interval. First, find the derivative: f'(x) = 3x² - 12x. Set f'(x) = 0 to find critical points: 3x(x - 4) = 0, so x=0 and x=4 are critical points. Now, evaluate the function at the critical points and the endpoints: f(-1) = -2, f(0) = 5, f(4) = -27, and f(5) = -20. The lowest value is -27, which occurs at x=4.
A) 2000 m²
B) 2500 m²
C) 3600 m²
D) 4000 m²
Correct Answer: B
This is an optimization problem. Let the dimensions of the rectangle be length (l) and width (w). The perimeter is P = 2l + 2w = 200, so l + w = 100, or l = 100 - w. The area is A = l * w. Substituting for l, we get the function A(w) = (100 - w) * w = 100w - w². To maximize the area, we take the derivative with respect to w: A'(w) = 100 - 2w. Set A'(w) = 0 to find the critical point: 100 - 2w = 0, which gives w = 50. If w = 50, then l = 100 - 50 = 50. The maximum area is 50 * 50 = 2500 m².
A) Only the values where the derivative is zero.
B) Only the values at the endpoints, f(a) and f(b).
C) The values at the endpoints and any critical points within the interval (a, b).
D) Only the values where the second derivative is zero.
Correct Answer: C
The Extreme Value Theorem guarantees that a continuous function on a closed interval will have an absolute maximum and minimum. These can occur either at the endpoints of the interval [a, b] or at critical points (where the derivative is zero or undefined) located within the open interval (a, b). Therefore, all these points must be evaluated in the original function to find the absolute extrema.
A) 72
B) 144
C) 0
D) 60
Correct Answer: A
Let the two numbers be x and y. We are given x + y = 12, and x ≥ 0, y ≥ 0. We want to minimize the function S = x² + y². From the first equation, we can write y = 12 - x. Substitute this into S: S(x) = x² + (12 - x)² = x² + 144 - 24x + x² = 2x² - 24x + 144. To find the minimum, take the derivative: S'(x) = 4x - 24. Set S'(x) = 0, which gives 4x = 24, so x = 6. When x = 6, y = 12 - 6 = 6. The minimum sum of squares is S = 6² + 6² = 36 + 36 = 72. We should also check the endpoints of the interval for x, which is [0, 12]. S(0) = 144, S(12) = 144. The minimum value is 72.
A) 2.12 cm
B) 5.42 cm
C) 10.0 cm
D) 8.77 cm
Correct Answer: B
We want to minimize surface area A = 2πr² + 2πrh, subject to the constraint V = πr²h = 1000. First, solve the volume constraint for h: h = 1000 / (πr²). Substitute this into the area formula: A(r) = 2πr² + 2πr(1000 / (πr²)) = 2πr² + 2000/r. To find the minimum, take the derivative with respect to r: A'(r) = 4πr - 2000/r². Set A'(r) = 0: 4πr = 2000/r², which simplifies to 4πr³ = 2000, or r³ = 500/π. Solving for r gives r = (500/π)^(1/3) ≈ 5.42 cm.
A) (-∞, -1)
B) (-1, 3)
C) (-∞, 3)
D) (-∞, -1) and (3, ∞)
Correct Answer: C
A function is decreasing where its derivative is negative. We need to find where g'(x) < 0. The critical points are x = 3 and x = -1. The term (x + 1)² is always non-negative. Therefore, the sign of g'(x) is determined by the sign of the term (x - 3). The expression (x - 3) is negative when x < 3. Thus, g'(x) is negative for all x < 3 (except at x=-1 where it is zero). This means the function g is decreasing on the interval (-∞, 3).