AP Calculus AB/BC · Unit 5 · Topic 5.10
Introduction to Optimization Problems
Translate a maximum or minimum context into an objective and a feasible constraint.
1. Topic Focus
Use derivatives to prove existence, classify extrema, analyze monotonicity and concavity, sketch graphs, and solve optimization problems.
This topic: Translate a maximum or minimum context into an objective and a feasible constraint.
2. Key Relationship
Read every symbol with its domain, direction, units, and hypotheses before applying the relationship.
3. Visual Connection
4. Worked Example
For a rectangle with fixed perimeter, express area using one side length.
Write the governing relationship first, carry out the algebra cleanly, and finish with a sentence that answers the mathematical question.
5. Concept Development
1. What optimization means
An optimization problem asks for the greatest or least possible value of a quantity under stated restrictions. Calculus enters only after the context has been translated into a function with a meaningful domain.
2. Separate the objective and the constraint
The objective is the quantity to maximize or minimize, such as area, volume, revenue, cost, distance, or time. A constraint relates the variables and limits the allowable choices, such as fixed perimeter, fixed volume, or a demand equation.
3. Define variables with units
Assign a symbol to every changing quantity and state its units. A labeled diagram often reveals repeated lengths, missing sides, right triangles, or surface pieces that prose can hide.
4. Translate the objective before differentiating
Write a formula for the requested quantity in its natural variables. For a rectangle, \(A=xy\); for travel, \(T=d_1/v_1+d_2/v_2\); for profit, \(P=R-C\). Do not differentiate a constraint when the question asks to optimize another quantity.
5. Reduce to one independent variable
Use the constraint to eliminate extra variables from the objective. Substitution should produce a single-variable function such as
Then verify that every occurrence of the eliminated variable has been replaced.
6. Build the feasible domain from the context
Lengths, times, and quantities are usually nonnegative. Geometry may impose stronger bounds: cutting squares of side \(x\) from a sheet of width \(W\) requires \(0 A degenerate shape may be mathematically included but physically excluded. It is often useful to analyze a continuous objective on a closed interval including zero-area or zero-volume endpoints, then interpret whether the maximizing interior point belongs to the original physical problem. Interior candidates occur where \(Q'(x)=0\) or where \(Q'\) is undefined while \(Q\) exists. On a closed interval, include both endpoints. Keep only candidates inside the feasible domain. A critical number alone is not an answer. Compare candidate values, use derivative signs, or apply an appropriate concavity argument. The Extreme Value Theorem guarantees existence when the objective is continuous on a closed, bounded interval. The optimizing input may represent only one dimension. Substitute it into the constraint to find remaining dimensions, then evaluate the objective. Report the requested dimensions and maximum or minimum value with units. Confirm dimensional consistency, feasible signs, and whether the model applies throughout the domain. A demand equation may be valid only over a stated price range, and a geometric formula may assume a particular orientation or symmetry.7. Decide whether endpoints are feasible
8. Generate every candidate
9. Justify a global result
10. Recover all requested quantities
11. Recognize common model families
12. Check the model and its assumptions
13. Common errors
6. Detailed Worked Example and Error Check
Example 1: Rectangle with fixed perimeter. A rectangle has perimeter \(40\) meters. Let the side lengths be \(x\) and \(y\). The objective is \(A=xy\), and the constraint \(2x+2y=40\) gives \(y=20-x\). Therefore
The interior candidate is \(x=10\); the constraint then gives \(y=10\). The endpoint models have zero area.
Example 2: Three-sided enclosure. A river supplies one side of a rectangular pen, and \(100\) meters of fencing forms the other three sides. If \(x\) is each perpendicular side and \(y\) is the parallel side, then \(2x+y=100\). The area model is
The derivative candidate is \(x=25\), corresponding to \(y=50\).
Example 3: Open-top box. Squares of side \(x\) are cut from a \(20\)-by-\(30\) sheet. The folded box has height \(x\), width \(20-2x\), and length \(30-2x\), so
Solving \(V'(x)=0\) produces \(x=(25\pm5\sqrt7)/3\); only the smaller value lies in the feasible interval.
Example 4: Fixed-volume container. An open cylindrical container has radius \(r\), height \(h\), and fixed volume \(V_0\). The objective surface area and constraint are
Substituting \(h=V_0/(\pi r^2)\) gives \(S(r)=\pi r^2+2V_0/r\) for \(r>0\). The model is now ready for calculus.
Example 5: Revenue model. Suppose demand is \(n(p)=500-2p\) items at price \(p\), for \(0\le p\le250\). Revenue is
The stationary price is \(p=125\). The stated demand interval prevents meaningless negative sales.
Example 6: Closest point. A point \((x,x^2)\) on the parabola \(y=x^2\) is to be closest to \((0,3)\). Minimizing distance is equivalent to minimizing
The candidates are \(x=0\) and \(x=\pm\sqrt{5/2}\). Squaring avoids a square root while preserving the location of the minimum because distance is nonnegative.
Example 7: Minimum travel time. An island is \(2\) miles offshore from point \(B\). A traveler begins \(6\) miles west of \(B\), runs at \(8\) mph, and swims at \(3\) mph. If \(x\) is the miles run toward \(B\), then
The two terms are times, not distances, and the domain represents all possible shoreline entry points between the start and \(B\).
7. AP Reasoning Routine
State theorem hypotheses, make sign charts on domain intervals, include endpoints when required, and connect derivative signs to function behavior.
- Identify the representation and requested quantity.
- State the rule or theorem and verify its conditions.
- Keep exact values until the final requested approximation.
- Interpret sign, units, interval, and context.
Construct the one-variable objective and feasible domain. Identify derivative candidates when requested.
(a) A rectangle has perimeter \(60\). Model its area using one side \(x\).
(b) A river borders one side of a rectangular pen made with \(120\) meters of fencing. Model its area.
(c) Squares of side \(x\) are cut from a \(16\)-by-\(24\) sheet to form an open box. Model its volume and domain.
(d) An open box with square base \(x\) and height \(h\) must have volume \(500\). Model its surface area using \(x\).
(e) Demand is \(n(p)=800-4p\). Model revenue over the range where demand is nonnegative.
(f) A cost model has fixed-plus-variable form \(C(x)=x^2+100/x\). State its meaningful domain and derivative candidates.
(g) Model the squared distance from \((0,2)\) to a point \((x,x^2)\) on \(y=x^2\).
(h) Explain why minimizing squared distance gives the same location as minimizing distance.
(i) Explain why endpoints must be checked in a closed-interval optimization problem.
(j) For a closed cylinder of radius \(r\), height \(h\), and fixed volume \(V_0\), identify the objective and constraint for minimizing material.
Check the solution
In part (a), \(2x+2y=60\) gives \(A(x)=x(30-x)\) on \([0,30]\). In part (b), \(2x+y=120\) gives \(A(x)=x(120-2x)\) on \([0,60]\). In part (c), \(V(x)=x(16-2x)(24-2x)\) on \([0,8]\); the open interval gives nondegenerate boxes. In part (d), \(x^2h=500\), so \(h=500/x^2\). The open-top surface area is \(S=x^2+4xh=x^2+2000/x\) for \(x>0\). In part (e), \(R(p)=p(800-4p)=800p-4p^2\) on \([0,200]\). In part (f), the domain is \(x>0\). Since \(C'(x)=2x-100/x^2\), the positive derivative candidate satisfies \(2x^3=100\), so \(x=\sqrt[3]{50}\). In part (g), \(D^2(x)=x^2+(x^2-2)^2\) for every real \(x\). In part (h), the square-root function is increasing on nonnegative inputs, so the distance and its square are minimized at the same input. In part (i), an absolute maximum or minimum can occur at an endpoint even when no derivative condition holds there; all feasible candidates must be compared. In part (j), the objective is the closed-cylinder surface area \(S=2\pi r^2+2\pi rh\), and the constraint is \(\pi r^2h=V_0\). Substitution gives \(S(r)=2\pi r^2+2V_0/r\) for \(r>0\).