A farmer has 200 feet of fencing and wants to enclose a rectangular pen along the side of a straight river. The side along the river needs no fence. What dimensions give the largest possible area, and what is that area?
Try it before reading on. Let the two sides perpendicular to the river have length x and the side parallel to the river have length y. Only three sides are fenced, so the fencing used is 2x + y = 200. The area is A = xy.
Here is the whole game of optimization in one question: you have two things you care about — the fence (fixed at 200) and the area (which you want as large as possible). One of them is the constraint; the other is the objective. Your job is to use the constraint to write the objective in terms of a single variable, then use calculus to find its maximum. Keep your guess. By the end of section (b) you will have the exact answer — and a justification good enough to earn full credit on the AP exam.
Optimization problems ask: what is the biggest or smallest value of some quantity, given a restriction? Maximize area, minimize cost, minimize material, minimize distance. Every one of them yields to the same six-step procedure. Memorize it as a routine — on the AP exam, the setup and the justification are where the points live.
The Optimization Procedure
- First Derivative Test: the objective has a maximum at x = c because f' changes from positive to negative there (minimum: negative to positive).
- Second Derivative Test: f'(c) = 0 and f''(c) < 0 ⟹ maximum; f''(c) > 0 ⟹ minimum.
- Candidates Test (closed-interval method): evaluate the objective at every critical point and at both endpoints; the largest value is the absolute max, the smallest is the absolute min.
A critical point is only a candidate. Finding
f'(c) = 0is step 4, not the answer. You must confirm it is a max or a min. Skipping the justification is the single most common way to lose a point on an AP optimization FRQ.
One fully worked problem (from the Opening Question)
A farmer has 200 ft of fencing for a rectangular pen along a river; the river side needs no fence. Maximize the area. [NO CALC]
Step 1 — Objective. We want to maximize area: A = xy, where x is each perpendicular side and y is the side parallel to the river.
Step 2 — Constraint → one variable. Three sides are fenced: 2x + y = 200. Solve for y:
y = 200 − 2xSubstitute into the objective:
A(x) = x(200 − 2x) = 200x − 2x²Now A is a function of the single variable x.
Step 3 — Domain. We need x > 0 and y = 200 − 2x > 0, so 0 < x < 100. (As a closed interval for the Candidates Test, 0 ≤ x ≤ 100, where the endpoints give zero area.)
Step 4 — Critical points.
A'(x) = 200 − 4x
A'(x) = 0 ⟹ x = 50Step 5 — Justify it's a maximum. A''(x) = −4 < 0 for all x, so by the Second Derivative Test, A has a relative maximum at x = 50. Because this is the only critical point on the interval and A''<0 everywhere (the parabola opens downward), it is the absolute maximum.
Step 6 — Answer with units. When x = 50: y = 200 − 2(50) = 100. The maximum area is
A = (50)(100) = 5000 square feet,achieved with the two perpendicular sides 50 ft each and the river-parallel side 100 ft.
Notice the pattern: half the fence (100 ft) ended up along the river, and the other half split between the two sides — a hallmark of these "three-sided" problems. But never rely on a remembered shortcut for credit; the derivative and the justification are what earn points.
Problem. A rancher has 2400 ft of fencing to build a rectangular field that is also divided into two equal parts by a fence parallel to one side. Find the dimensions that maximize the total enclosed area.
Strategy. Identify the objective (total area) and the constraint (total fence). The internal divider is an extra segment, so count the fencing carefully.
Solution. Let x be the length of the two sides crossed by the divider and y the length of the other pair (the divider is parallel to the y-sides, so it also has length... let the field be x wide and y long, with the divider parallel to the width x). Fence used = two lengths y + three widths x (two outer + one divider):
2y + 3x = 2400 ⟹ y = (2400 − 3x)/2 = 1200 − 1.5x
Objective: A = xy = x(1200 − 1.5x) = 1200x − 1.5x²
Domain: 0 < x < 800
A'(x) = 1200 − 3x = 0 ⟹ x = 400Justification. A''(x) = −3 < 0, so by the Second Derivative Test x = 400 gives the absolute maximum.
Answer. y = 1200 − 1.5(400) = 600. Dimensions: the divided direction x = 400 ft, the other y = 600 ft, giving maximum area A = (400)(600) = 240,000 square feet.
Problem. An open-top box is made from a 12 in × 12 in square sheet by cutting equal squares of side x from each corner and folding up the sides. What value of x maximizes the volume?
Strategy. The base after folding is (12 − 2x) by (12 − 2x), and the height is x. Volume is the objective; the domain comes from the geometry.
Solution.
V(x) = x(12 − 2x)², domain 0 < x < 6Differentiate (product + chain rule), then factor:
V'(x) = (12 − 2x)² + x·2(12 − 2x)(−2)
= (12 − 2x)[(12 − 2x) − 4x]
= (12 − 2x)(12 − 6x)
V'(x) = 0 ⟹ x = 6 (rejected, not in domain) or x = 2Justification (First Derivative Test). On (0, 2), take x = 1: V'(1) = (10)(6) > 0. On (2, 6), take x = 3: V'(3) = (6)(−6) < 0. Since V' changes from positive to negative at x = 2, V has a relative — and, being the only interior critical point, absolute — maximum there.
Answer. V(2) = 2(12 − 4)² = 2(8²) = 2(64) = 128 cubic inches, achieved with corner cuts of x = 2 in.
Problem. A cylindrical can must hold 355 cm³ (a standard soda can). Find the radius and height that minimize the surface area (top + bottom + side), and report the minimum surface area.
Strategy. Objective = total surface area. Constraint = fixed volume. Use the volume equation to eliminate h.
Solution.
Constraint (volume): πr²h = 355 ⟹ h = 355/(πr²)
Objective (surface): S = 2πr² + 2πrh
Substitute: S(r) = 2πr² + 2πr·355/(πr²) = 2πr² + 710/r, r > 0Differentiate and set to zero:
S'(r) = 4πr − 710/r² = 0 ⟹ 4πr³ = 710 ⟹ r³ = 710/(4π) ≈ 56.497
r ≈ 3.837 cmTI-84: solve 4πX − 710/X² = 0 with MATH → Solver, or graph Y₁ = 2πX² + 710/X and use 2nd → CALC → minimum.
Justification (Second Derivative Test). S''(r) = 4π + 1420/r³. For all r > 0 this is positive, so S is concave up and r ≈ 3.837 gives the absolute minimum.
Answer. h = 355/(π·3.837²) ≈ 7.675 cm (note h ≈ 2r, the classic least-material result). Minimum surface area:
S ≈ 2π(3.837)² + 710/3.837 ≈ 277.5 cm².Problem. Find the point(s) on the parabola y = x² closest to the point (0, 3).
Strategy. Minimize distance. A key trick: minimizing distance D is the same as minimizing D² (since D ≥ 0), and D² avoids the square root, making the derivative far cleaner.
Solution. A point on the curve is (x, x²). Then
D² = (x − 0)² + (x² − 3)² = x² + (x² − 3)², call it f(x)
f'(x) = 2x + 2(x² − 3)(2x) = 2x[1 + 2(x² − 3)] = 2x(2x² − 5)
f'(x) = 0 ⟹ x = 0 or x² = 5/2 ⟹ x = ±√(2.5) ≈ ±1.581Justification (Candidates / First Derivative Test). Compare candidates: f(0) = 0 + 9 = 9 ⟹ D = 3. At x = √2.5: f = 2.5 + (2.5 − 3)² = 2.5 + 0.25 = 2.75 ⟹ D = √2.75 ≈ 1.658. The smaller value occurs at x = ±√2.5, so those give the minimum distance (the critical point at x=0 is a local max of closeness — actually a relative maximum of f between the two minima).
Answer. The closest points are (√2.5, 2.5) and (−√2.5, 2.5), i.e. (±1.581, 2.5), each at distance √2.75 ≈ 1.658 units from (0, 3).
Optimizing the constraint instead of the objective.
What students do: differentiate the perimeter or volume equation. Why it's wrong: the constraint is fixed — its "derivative set to zero" is meaningless. Fix: clearly label which equation is the objective (the thing you max/min) and which is the constraint (the fixed restriction). Only the objective gets differentiated.
Never justifying max vs. min.
What students do: solve f'(x) = 0, find x = 50, and stop. Why it's wrong: a critical point could be a max, a min, or neither. The AP rubric awards a separate point for the justification. Fix: always finish with a First Derivative Test, Second Derivative Test, or Candidates Test, written in a complete sentence.
Ignoring the domain and endpoints.
What students do: report a critical point without checking whether it lies in the realistic domain, or forget that on a closed interval the extreme value may be at an endpoint. Fix: state the domain in step 3; on closed intervals, use the Candidates Test and evaluate the endpoints too.
Not answering the question asked.
What students do: the problem asks for the maximum area but they report x = 50. Why it's wrong: x is a dimension, not the area. Fix: reread the question. Plug the optimal variable back in to compute exactly what was requested.
Forgetting units (or using the wrong ones).
Fix: lengths in ft/cm, areas in ft²/cm², volumes in ft³/cm³. A bare number with no units can cost the final answer point.
x > 0, the minimum value of f(x) = x + 16/x isx-inch squares from the corners of a 6 in × 6 in sheet. The volume V(x) is maximized when x =V(x), confirms a critical point x = c gives a maximum volume?y = 2x + 1 closest to the origin has x-coordinate approximatelyV. Setting dS/dr = 0 for the surface area S = 2πr² + 2V/r gives the relationship between height and radiusShort Answer
A rectangular field bordering a river (no fence on the river side) is divided into two equal pens by a fence perpendicular to the river. Total fencing is 600 ft. Find the dimensions and the maximum total area. Show your setup, critical point, and justification.
(Justification) A box with a square base and open top must have volume 32,000 cm³. Find the dimensions that minimize the material used, and justify that your answer gives a minimum (not a maximum).
Find the dimensions of the rectangle of largest area that can be inscribed under the curve y = 9 − x² and above the x-axis, with its base on the x-axis (symmetric about the y-axis). Justify.
(Justification) Let f(x) = x³ − 12x on the closed interval [0, 4]. Find the absolute maximum and absolute minimum values of f, and justify using the Candidates Test.
A printed page has total area 384 in², with 1-inch margins on the sides and 1.5-inch margins top and bottom. Find the overall page dimensions that maximize the printed (text) area. Justify that your critical point gives a maximum.
> Section II, Part B style — NO CALCULATOR. Show all work. A correct answer with no supporting work or justification earns little credit.
FRQ. A manufacturer designs an open-top storage box with a square base. The box must have a volume of 32,000 cm³. Let x be the side length of the square base (in cm) and h the height (in cm). The amount of material used is the area of the base plus the four sides (there is no top).
(a) (2 points) Write the amount of material M as a function of x alone, and state the domain.
(b) (3 points) Find the value of x that minimizes the material used. Show the work that leads to your answer.
(c) (2 points) Justify that the value of x you found in part (b) gives the minimum amount of material.
(d) (2 points) Find the dimensions of the box and the minimum amount of material used. Include units. Interpret what the height h equals in relation to the base.
Total: 9 points
(a) Constraint (volume): x²h = 32,000, so h = 32,000/x².
Material = base + four sides = x² + 4xh. Substitute h:
M(x) = x² + 4x·(32,000/x²) = x² + 128,000/x, x > 0(1 pt for the correct material expression x² + 4xh; 1 pt for substituting the constraint to get M as a function of x with domain x > 0.)
(b) Differentiate and set equal to zero:
M'(x) = 2x − 128,000/x²
2x − 128,000/x² = 0 ⟹ 2x³ = 128,000 ⟹ x³ = 64,000 ⟹ x = 40So x = 40 cm.
(1 pt for M'(x); 1 pt for setting M'(x)=0; 1 pt for x = 40.)
(c) By the Second Derivative Test:
M''(x) = 2 + 256,000/x³
M''(40) = 2 + 256,000/64,000 = 2 + 4 = 6 > 0.Since M'(40) = 0 and M''(40) > 0, M is concave up at x = 40, so x = 40 gives a relative minimum. Because it is the only critical point for x > 0 and M''(x) > 0 for all x > 0, it is the absolute minimum.
(2 pts: 1 for a valid test correctly carried out, 1 for the explicit conclusion that it is a minimum. A First Derivative Test — M' goes from negative to positive at x = 40 — earns equal credit.)
(d) Dimensions: base x = 40 cm, height h = 32,000/40² = 32,000/1600 = 20 cm.
Minimum material:
M(40) = 40² + 128,000/40 = 1600 + 3200 = 4800 cm².Interpretation: the base side is 40 cm and the height is 20 cm, so h = x/2 — the optimal open-top box is half as tall as it is wide.
(1 pt for dimensions x = 40, h = 20 with units; 1 pt for M = 4800 cm² and the interpretation h = x/2.)
x = 40 and stop, assuming it "must" be the minimum. The AP rubric awards a dedicated point for the justification. Write a complete sentence naming the test and stating the conclusion: "Since M''(40) = 6 > 0, M is concave up at x = 40, so this is a minimum."2x² + 4xh. Students who forget "open-top" set up the wrong objective and lose the part (a) points and everything downstream. Read the geometry carefully.x = 40 without computing h and M forfeits part (d). Always return to what was asked and attach cm and cm².x³ = 64,000 ⟹ x = 40 (since 40³ = 64,000), not x = 400. Double-check cube roots.1. (C) 100. Numbers a and 20 − a; product P = a(20 − a) = 20a − a². P' = 20 − 2a = 0 ⟹ a = 10. P'' = −2 < 0 ⟹ max. Product = 10·10 = 100.
- (A) 40, (B) 80: products of non-optimal pairs. (D) 200: mistakes the sum-times-something or doubles incorrectly.
2. (B) 8. f'(x) = 1 − 16/x² = 0 ⟹ x² = 16 ⟹ x = 4 (positive branch). f(4) = 4 + 16/4 = 8. f''(x) = 32/x³ > 0 ⟹ min.
- (A) 4: reports x instead of f(x). (C) 16, (D) 32: arithmetic on the wrong quantity.
3. (D) 625. All four sides: 2x + 2y = 100 ⟹ y = 50 − x. A = x(50 − x), A' = 50 − 2x = 0 ⟹ x = 25 = y (a 25×25 square). A = 625 ft².
- (A) 2500: uses 50×50 (forgets the perimeter splits over four sides). (C), (B): non-optimal rectangles.
4. (A) 1. V = x(6 − 2x)², V' = (6 − 2x)(6 − 6x) = 0 ⟹ x = 3 (rejected) or x = 1. FDT: V' goes + → − at x = 1 ⟹ max.
- (D) 3: the rejected root (gives zero volume). (B), (C): miscalculations of the critical point.
5. (D) ≈ 3.41 cm. Let the side cost 1 per cm², top/bottom cost 2. Constraint πr²h = 500 ⟹ h = 500/(πr²). Cost C = 2·(2πr²) + 1·(2πrh) = 4πr² + 2πr·500/(πr²) = 4πr² + 1000/r. C' = 8πr − 1000/r² = 0 ⟹ r³ = 1000/(8π) = 125/π ≈ 39.79 ⟹ r ≈ 3.41 cm. C'' = 8π + 2000/r³ > 0 ⟹ min.
- (A) 2.71: uses r³ = 1000/(8π) but forgets the doubled top/bottom or mishandles the factor. (C) 4.30, (B) 5.42: cube-root or coefficient slips.
6. (C) V''(c) < 0. Second Derivative Test for a maximum.
- (D) V''(c) > 0: that confirms a minimum. (A): sign of the function, irrelevant. (B) V'(c) > 0: contradicts c being a critical point.
7. (B) 800 m². Three-sided: 2x + y = 80, A = x(80 − 2x) = 80x − 2x². A' = 80 − 4x = 0 ⟹ x = 20, y = 40. A'' = −4 < 0 ⟹ max. A = 20·40 = 800 m².
- (C) 3200: forgets the factor and squares incorrectly. (D), (A): non-optimal dimensions.
8. (A) ≈ −0.40. Minimize D² = x² + (2x + 1)². d/dx = 2x + 2(2x+1)(2) = 2x + 8x + 4 = 10x + 4 = 0 ⟹ x = −0.4. D²'' = 10 > 0 ⟹ min.
- (C) 0.20, (B) −0.20: sign/arithmetic slips. (D) 0.40: correct magnitude, wrong sign.
9. (A) h = 2r. S = 2πr² + 2V/r (with V = πr²h). S' = 4πr − 2V/r² = 0 ⟹ 4πr³ = 2V = 2πr²h ⟹ h = 2r. Classic minimal-material can.
- (D) h = r, (B), (C): algebra errors in solving for the ratio.
10. (B) ≈ 6.83 cm. Open-top cylinder: V = πr²h = 1000, h = 1000/(πr²). Surface (no top) S = πr² + 2πrh = πr² + 2000/r. S' = 2πr − 2000/r² = 0 ⟹ r³ = 1000/π ≈ 318.31 ⟹ r ≈ 6.83 cm. S'' = 2π + 4000/r³ > 0 ⟹ min.
- (D) 5.03: drops the factor of 2 on the side-area term. (A) 7.82, (C) 10.6: cube-root or coefficient slips.
11. River side unfenced; one internal divider perpendicular to the river (parallel to the x sides). Let x = each perpendicular segment, y = side parallel to river. Fence = 3x + y = 600 (two outer perpendicular + one divider + the parallel side). y = 600 − 3x. A = x(600 − 3x) = 600x − 3x². A' = 600 − 6x = 0 ⟹ x = 100, y = 300. Justification: A'' = −6 < 0 ⟹ absolute max. Answer: x = 100 ft, y = 300 ft; maximum total area A = 100·300 = 30,000 ft².
12. Open-top, square base, V = x²h = 32,000 ⟹ h = 32,000/x². Material M = x² + 4xh = x² + 128,000/x. M' = 2x − 128,000/x² = 0 ⟹ x³ = 64,000 ⟹ x = 40, h = 20. Justification (required): M'' = 2 + 256,000/x³; M''(40) = 2 + 4 = 6 > 0, so M is concave up ⟹ minimum. Answer: base 40 cm, height 20 cm, minimum material M = 1600 + 3200 = 4800 cm². (Same numbers as the FRQ — good cross-check.)
13. Rectangle base on x-axis, symmetric about y-axis, top corners on y = 9 − x². Width = 2x, height = 9 − x². Area A = 2x(9 − x²) = 18x − 2x³, domain 0 < x < 3. A' = 18 − 6x² = 0 ⟹ x² = 3 ⟹ x = √3 ≈ 1.732. Justification: A'' = −12x < 0 for x > 0 ⟹ max. Answer: width 2√3 ≈ 3.46, height 9 − 3 = 6; maximum area A = 2√3·6 = 12√3 ≈ 20.78 square units.
14. f(x) = x³ − 12x, [0, 4]. f'(x) = 3x² − 12 = 0 ⟹ x² = 4 ⟹ x = 2 (only x = 2 is in [0,4]; x = −2 is outside). Candidates Test — evaluate at the critical point and both endpoints:
f(0) = 0
f(2) = 8 − 24 = −16
f(4) = 64 − 48 = 16Justification: By the Candidates Test, comparing all candidates, the absolute maximum is f(4) = 16 (at x = 4) and the absolute minimum is f(2) = −16 (at x = 2).
15. Page w × H with total area wH = 384, so H = 384/w. Side margins 1 in each (remove 2 from width); top/bottom margins 1.5 in each (remove 3 from height). Printed area:
P = (w − 2)(H − 3) = (w − 2)(384/w − 3)
= 384 − 3w − 768/w + 6 = 390 − 3w − 768/w
P'(w) = −3 + 768/w² = 0 ⟹ w² = 256 ⟹ w = 16Justification: P''(w) = −1536/w³ < 0 for w > 0 ⟹ maximum. Answer: width w = 16 in, height H = 384/16 = 24 in. (Printed area P = (14)(21) = 294 in².) Overall page: 16 in × 24 in.
CalcIQ · Lesson 23 of 35 · Unit 5 — Analytical Applications of Differentiation. This lesson is independent study material aligned to the College Board AP Calculus AB Course and Exam Description; it is not endorsed by or affiliated with the College Board. AP® is a registered trademark of the College Board. All derivatives, critical points, and optimum values in this lesson have been recomputed and accuracy-reviewed; each optimum is confirmed as a maximum or minimum via the First Derivative Test, Second Derivative Test, or Candidates Test.