Optimization Problems in Calculus: Applications of the Derivative 4.5

4.5: Optimization Problems

Introduction to Optimization Problems

Optimization problems in calculus involve finding the maximum or minimum values of a function, often called the objective function, subject to given constraints. These problems are fundamental in applications ranging from geometry to economics and engineering.

• Objective Function: The function to be maximized or minimized.

• Constraint(s): Equations or inequalities that restrict the possible values of the variables.

• Method: Express the objective function in terms of a single variable using the constraint, then use calculus (derivatives) to find critical points.

Example 1: Minimizing a Linear Expression with a Product Constraint

Find positive numbers x and y such that $xy = 12$ and the sum $S = 2x + y$ is as small as possible.

• Constraint: $xy = 12$

• Objective Function: $S = 2x + y$

• Express $y$ in terms of $x$: $y = \frac{12}{x}$

• Substitute into $S$: $S(x) = 2x + \frac{12}{x}$

• Find minimum by setting derivative to zero: $S'(x) = 2 - \frac{12}{x^2} = 0$ $2 = \frac{12}{x^2} \implies x^2 = 6 \implies x = \sqrt{6}$ $y = \frac{12}{\sqrt{6}} = 2\sqrt{6}$

• Solution: $x = \sqrt{6}$, $y = 2\sqrt{6}$

Example Application: This type of problem models cost minimization under a fixed production constraint.

Example 2: Maximizing Area of a Rectangle with Fixed Perimeter

Find the dimensions of a rectangle with perimeter 100 m whose area is as large as possible.

• Constraint: $2x + 2y = 100$

• Objective Function: $A = xy$

• Express $y$ in terms of $x$: $y = 50 - x$

• Substitute into $A$: $A(x) = x(50 - x) = 50x - x^2$

• Find maximum by setting derivative to zero: $A'(x) = 50 - 2x = 0$ $x = 25$, $y = 25$

• Solution: The rectangle is a square with sides 25 m.

Example Application: Maximizing usable area for fencing projects.

Example 3: Maximizing Area with Multiple Constraints (Divided Field)

A farmer has 2400 ft of fencing to fence off a rectangular field, dividing it into three equal rectangular pieces. Find the dimensions of the field with largest area.

• Constraint: $6x + 4y = 2400$ (6 vertical segments, 4 horizontal)

• Objective Function: $A = 3xy$

• Express $x$ in terms of $y$: $6x = 2400 - 4y \implies x = \frac{1200 - 2y}{3}$

• Substitute into $A$: $A(y) = 3x y = (1200 - 2y)y = 1200y - 2y^2$

• Find maximum by setting derivative to zero: $A'(y) = 1200 - 4y = 0$ $y = 300$, $x = 200$

• Solution: Dimensions are $x = 200$ ft, $y = 300$ ft.

Example Application: Land management and agricultural planning.

Example 4: Minimizing Material for a Box with Fixed Volume

A box with a square base and open top must have a volume of 32 m3. Find the dimensions that minimize the amount of material used.

• Constraint: $x^2 y = 32$

• Objective Function (Surface Area): $A = x^2 + 4xy$

• Express $y$ in terms of $x$: $y = \frac{32}{x^2}$

• Substitute into $A$: $A(x) = x^2 + 4x \frac{32}{x^2} = x^2 + \frac{128}{x}$

• Find minimum by setting derivative to zero: $A'(x) = 2x - \frac{128}{x^2} = 0$ $2x = \frac{128}{x^2} \implies x^3 = 64 \implies x = 4$ $y = \frac{32}{16} = 2$

• Solution: Base sides $x = 4$ m, height $y = 2$ m.

Example Application: Packaging design to minimize cost.

Example 5: Maximizing Volume of a Box with Linear Constraint

Suppose an airline policy states that all baggage must be box-shaped with the sum of length, width, and height not exceeding 72 in. What are the dimensions of a square-based box with the greatest volume?

• Constraint: $2x + y = 72$ (square base: length = width = x)

• Objective Function: $V = x^2 y$

• Express $y$ in terms of $x$: $y = 72 - 2x$

• Substitute into $V$: $V(x) = x^2 (72 - 2x) = 72x^2 - 2x^3$

• Find maximum by setting derivative to zero: $V'(x) = 144x - 6x^2 = 0$ $144x = 6x^2 \implies x = 24$ $y = 72 - 2 \times 24 = 24$

• Solution: Base sides $x = 24$ in, height $y = 24$ in.

Example Application: Maximizing luggage capacity under airline restrictions.

General Steps for Solving Optimization Problems

1. Identify the objective function to be maximized or minimized.

2. Write down the constraint(s) relating the variables.

3. Express the objective function in terms of a single variable using the constraint.

4. Differentiate the objective function and set the derivative equal to zero to find critical points.

5. Check endpoints and/or use the second derivative test to confirm maximum or minimum.

6. Interpret the solution in the context of the problem.

Summary Table: Optimization Problem Types

Problem Type	Objective Function	Constraint	Method
Minimize sum	$S = 2x + y$	$xy = c$	Express $y$ in terms of $x$, minimize $S(x)$
Maximize area	$A = xy$	$2x + 2y = P$	Express $y$ in terms of $x$, maximize $A(x)$
Maximize area (divided field)	$A = 3xy$	$6x + 4y = F$	Express $x$ in terms of $y$, maximize $A(y)$
Minimize material (box)	$A = x^2 + 4xy$	$x^2 y = V$	Express $y$ in terms of $x$, minimize $A(x)$
Maximize volume (box)	$V = x^2 y$	$2x + y = S$	Express $y$ in terms of $x$, maximize $V(x)$

Additional info: These examples illustrate classic applications of derivatives in solving real-world optimization problems, a key topic in Calculus Chapter 4: Applications of the Derivative.