Introduction

This topic introduces optimization—using calculus to find maximum and minimum values in practical situations. The essential knowledge tells us that the derivative can be used to solve optimization problems; that is, finding a minimum or maximum value of a function on a given interval. This foundational understanding prepares us for solving real-world optimization applications.

What Is Optimization?

The Basic Concept

Optimization means finding the best possible solution to a problem—usually finding a maximum or minimum value.

Common optimization goals:

Why Calculus Is Needed

In many real-world problems, we need to find the optimal value subject to constraints. Calculus allows us to:

The Basic Optimization Process

Step-by-Step Framework

Step 1: Identify what quantity needs to be maximized or minimized

Step 2: Express this quantity as a function of one variable

Step 3: Determine the domain (realistic constraints)

Step 4: Find critical points using the derivative

Step 5: Evaluate the function at critical points and endpoints

Step 6: Identify the maximum or minimum value

Example 1:

Find two positive numbers whose sum is $20$ and whose product is maximum.

Step 1: Maximize the product $P$

Step 2: Let the numbers be $x$ and $y$ where $x+y=20$

Express $P$ in terms of one variable:

$y=20-x$

$P(x)=x\cdot y=x(20-x)=20x-x^2$

Step 3: Domain: $0<x<20$ (both numbers must be positive)

Step 4: Find critical points:

$P^{\prime }(x)=20-2x$

$20-2x=0$

$x=10$

Step 5: Since this is an open interval, check the critical point and behavior at endpoints: $P(10)=10(10)=100$

As $x\to 0^{+}$ or $x\to 20^{-}$, $P\to 0$

Step 6: Maximum product is $100$ when both numbers are $10$.

Simple Geometric Optimization

Area Problems

Example 2:

A rectangle has a perimeter of 40 meters. What dimensions maximize the area?

Let length = $x$ and width = $y$

Constraint: $2x+2y=40$, so $y=20-x$

Area to maximize: $A(x)=xy=x(20-x)=20x-x^2$

Domain: $0<x<20$

Find critical points:

$A^{\prime }(x)=20-2x=0$

$x=10$

Therefore $y=10$

Maximum area: $A(10)=100$ square meters (when rectangle is a square)

Volume Problems

Example 3:

An open-top box is made from a 12 by 12 inch square of cardboard by cutting squares of side length $x$ from each corner and folding up the sides. What value of $x$ maximizes the volume?

Volume function: $V(x)=x(12-2x)(12-2x)=x(12-2x{)}^2$

Domain: $0<x<6$ (can't cut more than half the side)

Find critical points:

$V(x)=x(144-48x+4x^2)=144x-48x^2+4x^3$

$V^{\prime }(x)=144-96x+12x^2=12(12-8x+x^2)=12(x-2)(x-6)$

$V^{\prime }(x)=0$ when $x=2$ or $x=6$

Only $x=2$ is in the domain $(0,6)$.

Check: $V(2)=2(8)(8)=128$ cubic inches

Maximum volume is $128$ cubic inches when $x=2$ inches.

Optimization on Closed Intervals

Using the Candidates Test

When optimizing on a closed interval, use the Candidates Test from Topic 5.5.

Example 4:

Find the maximum value of $f(x)=3x^4-4x^3$ on $[-1,2]$.

Find critical points: $f^{\prime }(x)=12x^3-12x^2=12x^2(x-1)$ $x=0$ and $x=1$

Evaluate at critical points and endpoints:

$f(-1)=3+4=7$

$f(0)=0$

$f(1)=3-4=-1$

$f(2)=48-32=16$

Maximum value: $16$ at $x=2$

Minimum value: $-1$ at $x=1$

Identifying the Objective Function

Translating Words to Mathematics

A critical skill is identifying what function to optimize.

Example 5:

A farmer has 200 feet of fencing and wants to enclose a rectangular area. What is the objective function for the area?

Let $x$ = length and $y$ = width

Constraint: $2x+2y=200$, so $y=100-x$

Objective function: $A(x)=x(100-x)=100x-x^2$

This function will be maximized in the next topic using the optimization process.

Common Objective Functions

Area of rectangle: $A=xy$ (with constraint on perimeter)

Volume of box: $V=xyz$ (with constraint on surface area)

Distance: $D=\sqrt{(x_2-x_1{)}^2+(y_2-y_1{)}^2}$

Cost: $C=(\text{price per unit})\times (\text{number of units})$

Revenue: $R=(\text{price})\times (\text{quantity sold})$

Practice Section