AP® Calculus BC review sheet from Aim for Five (aimforfive.com/calc-bc/units/5/5-11)
Unit 5 · Topic 5.11
5.11 Solving Optimization Problems
With the objective function and domain set up, find the critical points, then prove which one gives the max or min, using the Candidates Test or a derivative test. Finish by answering the actual question, in context and with units.
Key terms
- critical point
- Candidates Test
- First Derivative Test
- maximize
- minimize
- context
Solve, then justify
After the setup from 5.10, the work follows a fixed pattern:
- Differentiate the one-variable objective function.
- Find the critical points in the domain.
- Justify the max or min: on a closed interval, compare candidates (critical points and endpoints). On an open or infinite interval, use the First or Second Derivative Test and note that there's only one critical point.
- Answer the question asked: the location, the optimal value, or both, with units.
Which justification to use
A critical point alone doesn't prove anything. If the domain is a closed interval, the Candidates Test is the cleanest proof. If it's open, like r > 0 for a can's radius, then show the critical point is a relative min or max with a derivative test, then say it's the only critical point on the interval, so it's the absolute min or max.
On calculator-active questions, solving f′(x) = 0 on the calculator is fine, but you still need the justification in words.
Endpoints can win
If a constraint narrows the domain, the best value may be at an endpoint where f′ isn't 0. Always compare the endpoints when the interval is closed. If the derivative has one sign on the whole domain, the function is monotonic (always increasing or always decreasing), and the optimum is at an end.
Calculator-active optimization
When the objective function is messy, like a distance from a point to a curve such as y = eˣ, finding where the derivative is 0 by hand may be impossible. On calculator questions, set up the derivative yourself, then use the calculator to solve f′(x) = 0 or to find the zero of the graph of f′.
Store the solution as a variable, then evaluate the objective with the stored value. Rounding a critical point to three decimals before plugging it in can change your final answer's third decimal place, which can cost the point.
Interpreting the answer
Minimum and maximum values mean something in context: the cheapest cost, the largest area, the shortest time. Read the question to see whether it wants the x-value (“what dimensions?”), the optimal value (“what is the maximum volume?”) or both, and include units. Round to three decimal places on calculator answers.
Worked examples
Try each one yourself first, then open the solution.
- Example 1
Finishing the open box
Maximize V(x) = x(12 − 2x)² on [0, 6] from 5.10, and give the box's dimensions.
Show the solutionHide the solution
- Step 1: Product and chain rules: V′(x) = (12 − 2x)² + x·2(12 − 2x)(−2) = (12 − 2x)[(12 − 2x) − 4x] = (12 − 2x)(12 − 6x).
- Step 2: V′(x) = 0 at x = 6 (an endpoint) and x = 2.
- Step 3: Candidates: V(0) = 0, V(2) = 2·8² = 128, V(6) = 0.
- Step 4: The largest is 128 at x = 2. The base is 12 − 4 = 8 inches on a side, and the height is 2 inches.
Answer: The maximum volume is 128 cubic inches, with a base of 8 in by 8 in and a height of 2 in.
- Example 2Calculator allowed
Least material for a can
A closed cylindrical can must hold 500 cm³. Find the radius and height that use the least total surface area.
Show the solutionHide the solution
- Step 1: Constraint: πr²h = 500, so h = 500/(πr²). Objective: S = 2πr² + 2πrh.
- Step 2: Substitute: S(r) = 2πr² + 2πr·500/(πr²) = 2πr² + 1000/r, for r > 0 (an open interval).
- Step 3: S′(r) = 4πr − 1000/r² = 0 gives r³ = 250/π, so r = ∛(250/π) ≈ 4.301 cm.
- Step 4: S″(r) = 4π + 2000/r³ > 0 for r > 0, so this is a relative minimum. It's the only critical point on r > 0, so it's the absolute minimum.
- Step 5: h = 500/(πr²) ≈ 8.603 cm, which is exactly 2r. The surface area is about 348.734 cm².
Answer: r ≈ 4.301 cm and h ≈ 8.603 cm (the height equals the diameter), using about 348.734 cm² of material.
- Example 3
Trap: the critical point is outside the domain
The farmer from 5.10 still has 600 m of fence along a river, but the side parallel to the river can be at most 200 m long. Find the largest possible area.
Show the solutionHide the solution
- Step 1: Same setup: A(x) = x(600 − 2x), with y = 600 − 2x.
- Step 2: New domain: y ≤ 200 means 600 − 2x ≤ 200, so x ≥ 200. Also x ≤ 300. Domain: [200, 300].
- Step 3: A′(x) = 600 − 4x = 0 at x = 150, which is outside [200, 300]. Using it would give a field with y = 300, breaking the rule.
- Step 4: On [200, 300], A′(x) < 0, so A is decreasing, and the max is at the left endpoint: A(200) = 200·200 = 40,000.
Answer: The largest area is 40,000 m², with x = 200 m and y = 200 m.
Common mistakes
- Stopping at the critical point without justifying that it's a max or min.
- Answering with x when the question asks for the maximum value, or forgetting units.
- Using a critical point that lies outside the domain.
On the exam
- Justification is what separates full credit from partial credit. Use the Candidates Test on closed intervals and a one-critical-point argument on open ones.
- Calculator-active optimization questions expect you to set up the derivative yourself and then use the calculator to solve. Store intermediate values instead of rounding them.
Connected topics
Videos
Check yourself
4 questions on 5.11 Solving Optimization Problems. Pick an answer to see if you got it, and why.
What is the minimum distance between the origin and a point on the graph of y = eˣ?
An open-top box with a square base must hold 32 cubic feet. Which dimensions use the least material for the base and four sides?
A farmer is building a rectangular pen. Fencing for the north side costs $10 per foot, and fencing for the other three sides costs $5 per foot. The farmer will spend exactly $600 on fencing. Let x be the length, in feet, of the north side, and let y be the length, in feet, of the east side.
Invented scenario
What is the maximum possible area of the pen?
Let P be a point on the graph of y = √x, and let D be the distance from P to the point (3, 0).
Described function
What is the minimum value of D ?
0 of 4 answered