AP Calculus AB: Optimization (Drill 1)

Question 1. Find the absolute maximum value of \( f(x) = x^3 - 3x^2 + 1 \) on the interval \( [0, 3] \).

• A) \( 1 \) ✓
• B) \( -1 \)
• C) \( -3 \)
• D) \( 3 \)

Explanation: Choice A is correct. Apply the Candidates Test: evaluate \( f \) at all interior critical points and at both endpoints. \( f'(x) = 3x^2 - 6x = 3x(x - 2) = 0 \) gives interior critical point \( x = 2 \) (with \( x = 0 \) itself an endpoint). Evaluating all candidates: \( f(0) = 1 \), \( f(2) = 8 - 12 + 1 = -3 \), \( f(3) = 27 - 27 + 1 = 1 \). The largest value is \( 1 \), and notably it occurs at both endpoints \( x = 0 \) and \( x = 3 \), a function can attain its absolute maximum at more than one point. Choice B is off because \( -1 \) does not appear among the candidate values and results from an arithmetic error. Choice C doesn't fit: \( -3 \) is the absolute minimum, occurring at the interior critical point \( x = 2 \), a student who evaluates only the critical point and ignores the endpoints would make this error. Choice D is wrong because \( 3 \) is the x-value of the right endpoint, not the function’s value there, a student who confuses the input with the output would choose this.

Question 2. Which of the following lists all critical points of \( f(x) = x^4 - 8x^2 + 3 \)?

• A) \( x = 2 \) only
• B) \( x = 0 \) only
• C) \( x = -2 \), \( x = 0 \), and \( x = 2 \) ✓
• D) \( x = -2 \) and \( x = 2 \) only

Explanation: Choice C is correct. \( f'(x) = 4x^3 - 16x = 4x(x^2 - 4) = 4x(x-2)(x+2) \). Setting \( f'(x) = 0 \) gives \( x = 0 \), \( x = 2 \), and \( x = -2 \). Since \( f \) is differentiable everywhere, all three are critical points. Choice A is incorrect because it omits \( x = 0 \) and \( x = -2 \), likely from only solving \( x^2 = 4 \) and ignoring the factor \( 4x \). Choice B is incorrect because it omits the solutions from \( x^2 - 4 = 0 \), likely from only setting \( 4x = 0 \). Choice D is incorrect because it omits \( x = 0 \); a common error is failing to factor completely and missing the root at \( x = 0 \).

Question 3. A farmer has 200 feet of fencing and wants to enclose a rectangular area using a straight barn wall as one side (so fencing is needed for only three sides). What dimensions maximize the enclosed area?

• A) Length parallel to barn = 100 ft, width = 50 ft ✓
• B) Length parallel to barn = 50 ft, width = 75 ft
• C) Length parallel to barn = 80 ft, width = 60 ft
• D) Length parallel to barn = 120 ft, width = 40 ft

Explanation: Choice A is correct. Let \( L \) = length parallel to the barn and \( w \) = width (two sides of fencing needed). The constraint is \( L + 2w = 200 \), so \( L = 200 - 2w \). Area: \( A(w) = Lw = (200 - 2w)w = 200w - 2w^2 \). Setting \( A'(w) = 200 - 4w = 0 \) gives \( w = 50 \), and \( L = 200 - 100 = 100 \). Since \( A''(w) = -4 < 0 \), this is a maximum. Choice B is off because this setup likely stems from using \( L + w = 200 \) (forgetting that two widths are needed), which gives the wrong constraint. Choice C doesn't fit: these dimensions do not satisfy the constraint \( L + 2w = 200 \) (\( 80 + 120 \neq 200 \)) and appear to result from a setup error. Choice D is wrong because while \( 120 + 2(40) = 200 \) satisfies the constraint, this is not the critical point, a student may have guessed a 60/40 split of the total fencing rather than applying calculus.

Question 4. A cylindrical can is to be manufactured with a fixed volume \( V \). The material for the top and bottom costs twice as much per square unit as the material for the lateral (side) surface. Using \( h = \dfrac{V}{\pi r^2} \) to eliminate \( h \), which of the following correctly represents the total cost \( C \) as a function of radius \( r \) (taking the cost per unit area of the lateral surface as \( 1 \))?

• A) \( C = 2\pi r^2 + \dfrac{2V}{r} \)
• B) \( C = 4\pi r^2 + \dfrac{2V}{r} \) ✓
• C) \( C = 2\pi r^2 + \dfrac{V}{r} \)
• D) \( C = 4\pi r^2 + \dfrac{V}{\pi r} \)

Explanation: Choice B is correct. The top and bottom each have area \( \pi r^2 \) and cost 2 per unit area, contributing \( 2 \times 2\pi r^2 = 4\pi r^2 \). The lateral surface has area \( 2\pi r h = 2\pi r \cdot \dfrac{V}{\pi r^2} = \dfrac{2V}{r} \) and costs 1 per unit area, contributing \( \dfrac{2V}{r} \). Total: \( C = 4\pi r^2 + \dfrac{2V}{r} \). Choice A is incorrect because it uses \( 2\pi r^2 \) instead of \( 4\pi r^2 \) for the top and bottom, the student applied the area formula correctly but forgot to double the cost per unit area for those surfaces. Choice C is incorrect on both terms; it appears the student used cost factor 1 for the top and bottom and also made an error substituting \( h \) into the lateral area. Choice D is incorrect in the lateral surface term; the \( \pi \) in the denominator results from not fully simplifying \( 2\pi r \cdot \dfrac{V}{\pi r^2} \), leaving an extra \( \pi \) incorrectly in the denominator.

Question 5. A 10-foot wire is cut into two pieces. One piece is bent into a square and the other into a circle. If \( x \) represents the length of wire used for the square (so \( 10 - x \) is used for the circle), where \( 0 \leq x \leq 10 \), which value of \( x \) minimizes the combined area of the two shapes?

• A) \( x = \dfrac{40}{4 + \pi} \) ✓
• B) \( x = 5 \)
• C) \( x = 0 \) (all wire to the circle)
• D) \( x = 10 \) (all wire to the square)

Explanation: Choice A is correct. The side of the square is \( \dfrac{x}{4} \), so its area is \( \dfrac{x^2}{16} \). The radius of the circle is \( \dfrac{10-x}{2\pi} \), so its area is \( \dfrac{(10-x)^2}{4\pi} \). Total area: \( A(x) = \dfrac{x^2}{16} + \dfrac{(10-x)^2}{4\pi} \). Setting \( A'(x) = \dfrac{x}{8} - \dfrac{10-x}{2\pi} = 0 \) and solving: \( 2\pi x = 8(10 - x) \Rightarrow x(2\pi + 8) = 80 \Rightarrow x = \dfrac{40}{4 + \pi} \approx 5.60 \). Since \( A''(x) = \dfrac{1}{8} + \dfrac{1}{2\pi} > 0 \), this critical point is a minimum. Checking all candidates: \( A\!\left(\dfrac{40}{4+\pi}\right) \approx 3.50 \), \( A(0) = \dfrac{100}{4\pi} \approx 7.96 \), \( A(10) = \dfrac{100}{16} = 6.25 \). The minimum is at the interior critical point. Note: if the question asked for the maximum combined area, the answer would be \( x = 0 \) (all wire to the circle), since \( 7.96 > 6.25 \), the maximum always occurs at an endpoint. Choice B doesn't fit: \( x = 5 \) is a natural guess but is not the calculus-derived minimum. Choice C is incorrect for the minimum question; using all wire for the circle yields the largest combined area (\( \approx 7.96 \)). Choice D is wrong because using all wire for the square gives area \( 6.25 \), which is larger than the value at the critical point.