5.10 Introduction to Optimization Problems

Learning objectives

—FUN-4.B—Calculate minimum and maximum values in applied contexts or analysis of functions• FUN-4.B Calculate minimum and maximum values in applied contexts or analysis of functions.• FUN-4.B.1 The derivative can be used to solve optimization problems; that is, finding a minimum or maximum value of a function on a given interval.• Enduring understanding FUN-4: A function’s derivative can be used to understand some behaviors of the function.

An optimization problem asks for the greatest or least value of an objective quantity subject to constraints. Calculus begins only after the objective has been written as a function of one variable and its feasible interval has been identified.

Name the quantity to maximize or minimize.\n2. Write an objective equation and a separate constraint.\n3. Use the constraint to express the objective in one variable.\n4. Determine the feasible domain from the context.\n5. Find interior critical points and compare all valid candidates, including endpoints when they belong to the interval.\n6. Answer with the requested quantity, units, and a contextual interpretation.

For a rectangle with fixed perimeter $P>0$ , let its sides be $x$ and $y$ . The constraint $2x+2y=P$ gives $y=P/2-x$ . Therefore $A(x)=x\left(\frac{P}{2}-x\right),\qquad 0\le x\le\frac{P}{2}.$ Differentiate: $A'(x)=P/2-2x$ , so the interior candidate is $x=P/4$ . Then $y=P/4$ . The endpoint areas are $0$ , while $A(P/4)=P^2/16$ , so the maximum area is $P^2/16$ square units and occurs for a square.

Solving $A'(x)=0$ produces a candidate input, not automatically the requested optimum. Check that it is feasible, compare it with every required endpoint or other critical point, and distinguish the optimizing dimensions from the maximum or minimum value itself.

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—5.11 Solving Optimization ProblemsRead