AP® Calculus BC review sheet from Aim for Five (aimforfive.com/calc-bc/units/5/5-10)
Unit 5 · Topic 5.10
5.10 Introduction to Optimization Problems
Optimization problems ask for the largest or smallest possible value of something: area, volume, cost, distance. The hardest part is the setup: write one function for the quantity to optimize, use the constraint to reduce it to a single variable, and find the realistic domain.
Key terms
- optimization
- objective function
- constraint
- domain
Two kinds of equations
Every optimization problem gives you two kinds of information:
- Objective function: the formula for the quantity you want to make as large or small as possible, like A = xy for area.
- Constraint: a fact that ties the variables together, like “600 meters of fence” giving 2x + y = 600.
Reduce to one variable
The objective usually starts with two variables. Solve the constraint for one of them and substitute. With A = xy and y = 600 − 2x, the area becomes A(x) = x(600 − 2x), a function of x alone. Now you can use derivatives.
Pick the variable that makes the algebra easiest. Often that's the one that appears without a square or root in the constraint.
Find the domain
Real situations limit the variable. Lengths can't be negative, and they can't be so large that another length becomes negative. In the fence problem, x ≥ 0 and y = 600 − 2x ≥ 0, so 0 ≤ x ≤ 300.
The domain decides your method later. On a closed interval, you can use the Candidates Test. On an open or infinite interval, you'll use a derivative test with the one-critical-point idea (5.7).
A setup checklist
Use these steps before you differentiate anything:
- Draw a picture and label everything with variables.
- Write the objective function in words first, then as a formula.
- Write the constraint as an equation.
- Substitute to get the objective in one variable.
- State the domain.
Common objective functions
Many problems reuse the same formulas. Have these ready:
- Rectangle: area A = lw, perimeter P = 2l + 2w.
- Box with a square base: volume V = s²h, surface area (open top) s² + 4sh.
- Cylinder: volume V = πr²h, total surface area 2πr² + 2πrh.
- Distance between (x, y) and (a, b): √((x − a)² + (y − b)²).
- Profit = revenue − cost, where revenue = (price)(quantity).
A useful shortcut for distance
To minimize a distance D = √(…), you can minimize D² instead. Since √ is increasing, D and D² are smallest at the same x, and D² has no square root to differentiate. Just remember to take the square root at the end if the question asks for the distance.
Worked examples
Try each one yourself first, then open the solution.
- Example 1
Fence along a river
A farmer has 600 m of fence to enclose a rectangular field along a straight river. No fence is needed along the river. Write the area as a function of one variable, give its domain, and find the largest area.
Show the solutionHide the solution
- Step 1: Let x be each side perpendicular to the river and y the side parallel to it. Constraint: 2x + y = 600, so y = 600 − 2x.
- Step 2: Objective: A = xy = x(600 − 2x) = 600x − 2x².
- Step 3: Domain: x ≥ 0 and 600 − 2x ≥ 0, so 0 ≤ x ≤ 300.
- Step 4: A′(x) = 600 − 4x = 0 at x = 150. Candidates: A(0) = 0, A(150) = 45,000, A(300) = 0.
Answer: A(x) = x(600 − 2x) on [0, 300]. The largest area is 45,000 m², with x = 150 m and y = 300 m.
- Example 2
Setting up an open box
Equal squares of side x are cut from the corners of a 12-inch by 12-inch sheet, and the sides are folded up to make an open box. Write the volume as a function of x and give its domain.
Show the solutionHide the solution
- Step 1: Cutting x from each side leaves a square base of side 12 − 2x. The height is x.
- Step 2: V = (base area)(height) = (12 − 2x)²·x.
- Step 3: Domain: x ≥ 0 and 12 − 2x ≥ 0, so 0 ≤ x ≤ 6. (At the ends, the volume is 0.)
Answer: V(x) = x(12 − 2x)² on [0, 6]. (5.11 finishes this problem.)
- Example 3
Trap: minimizing distance with a square root
Find the point on the curve y = √x that is closest to the point (3, 0).
Show the solutionHide the solution
- Step 1: A point on the curve is (x, √x). Its distance to (3, 0) is D = √((x − 3)² + (√x − 0)²).
- Step 2: Minimize D² instead: D² = (x − 3)² + x = x² − 5x + 9. Domain: x ≥ 0, since √x needs it.
- Step 3: d(D²)/dx = 2x − 5 = 0 at x = 5/2. The second derivative is 2 > 0, and it's the only critical point on [0, ∞), so it gives the absolute minimum.
- Step 4: At x = 5/2, D² = 25/4 − 25/2 + 9 = 11/4, so D = √11/2 ≈ 1.658. (The endpoint x = 0 gives D = 3, which is larger.)
- Step 5: A common slip is minimizing (x − 3)² + √x, forgetting that the y-difference is squared too: (√x)² = x.
Answer: The closest point is (5/2, √(5/2)), at a distance of √11/2 ≈ 1.658.
Common mistakes
- Differentiating the objective while it still has two variables.
- Ignoring the domain, which can hide the true max or min at an endpoint.
- Optimizing the constraint instead of the objective, like maximizing the fence length instead of the area.
On the exam
- Optimization in free response usually gives the objective function directly or asks for one setup step. Multiple-choice questions are more likely to ask for the full setup.
- Always find the domain. It tells you whether to use the Candidates Test or a one-critical-point argument.
Connected topics
Videos
Check yourself
4 questions on 5.10 Introduction to Optimization Problems. Pick an answer to see if you got it, and why.
A closed cylindrical can (with a top and a bottom) must hold 500 cubic centimeters. Which of the following gives the total surface area S of the can, in square centimeters, as a function of its radius r?
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
Which of the following gives the area A of the pen as a function of x, along with a domain that makes sense in context?
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
Let x be the x-coordinate of P. Since minimizing D² gives the same point as minimizing D, which of the following functions should be minimized, for x ≥ 0, to find the point P closest to (3, 0) ?
0 of 4 answered