Optimization in Business Calculus: Methods and Applications

§3.5 Optimization: Business, Economics, and General Applications

This section introduces optimization problems commonly encountered in business calculus, focusing on maximizing or minimizing quantities such as revenue, profit, and cost. The methods and examples provided are essential for understanding how calculus is applied to real-world business and economic scenarios.

Business Formulae

Key Definitions and Formulas

• Revenue (R): The total income from sales, calculated as the price per item multiplied by the number of items sold.

• Price-Demand Function (p(x)): A function expressing the price at which x items can be sold. Typically, as the number of items sold increases, the price per item decreases.

• Revenue Function: If p(x) is the price-demand function, then revenue is given by: $R(x) = x p(x)$

• Cost Function (C(x)): The total cost of producing x items.

• Profit Function (P(x)): The net profit is the difference between revenue and cost: $P(x) = R(x) - C(x)$

Example Application: If a company sells x units at price p(x) and incurs cost C(x), then maximizing profit or revenue involves finding the value of x that yields the highest P(x) or R(x).

Optimization Example #1: Maximizing Revenue and Profit

Problem Statement

A company manufactures and sells x digital cameras per week. The price-demand and cost equations are:

• Price-demand: $p = 400 - 0.4x$

• Cost: $C(x) = 2,000 + 120x$

1. (A) Revenue Maximization: What price should the company charge, and how many cameras should be produced to maximize weekly revenue? What is the maximum revenue?

2. (B) Profit Maximization: What is the maximum weekly profit? What price should the company charge, and how many cameras should be produced to realize the maximum profit?

Solutions

• (A) Maximum Revenue: Price = $200, Number of cameras = 500, Maximum revenue = $100,000

• (B) Maximum Profit: Price = $280, Number of cameras = 300, Maximum profit = $34,000

Example Explanation: To solve, express revenue and profit as functions of x, find their critical points using calculus, and evaluate at those points.

General Method of Attack for Optimization Problems

Step-by-Step Problem Solving Strategy

1. Identify the quantity to be optimized. (e.g., maximize profit, minimize cost, maximize area, etc.)

2. Find a formula for the quantity. (Write an equation for the quantity in terms of relevant variables.)

3. Express the quantity in terms of one variable. (Use constraints to eliminate extra variables, so the function is Q(x).)

4. Find the critical numbers of Q(x). (Set the derivative $Q'(x) = 0$ and solve for x.)

5. Verify that the critical number produces a maximum or minimum. (Use the second derivative test or analyze endpoints if necessary.)

6. Answer the original question. (Interpret the result in the context of the problem.)

Additional info: This method is general and applies to a wide range of optimization problems in business, economics, and geometry.

Additional Optimization Examples

Example #2: Maximizing Product with a Fixed Sum

Problem: Find two numbers whose sum is 21 and whose product is a maximum.

• Solution: The numbers are 10.5 and 10.5.

Explanation: Let the numbers be x and 21 - x. The product is $P(x) = x(21 - x)$. Maximizing P(x) yields x = 10.5.

Example #3: Minimizing Sum with a Fixed Product

Problem: Find two positive numbers whose product is 21 and whose sum is a minimum.

• Solution: The numbers are both $\sqrt{21}$.

Explanation: Let the numbers be x and 21/x. The sum is $S(x) = x + \frac{21}{x}$. Minimizing S(x) yields $x = \sqrt{21}$.

Example #4: Minimizing Perimeter with Fixed Area (Rectangle)

Problem: Find the dimensions of a rectangle with an area of 108 square feet that has the minimum perimeter.

• Solution: Dimensions are $6\sqrt{3}$ ft by $6\sqrt{3}$ ft.

Explanation: For area A = xy = 108, perimeter P = 2x + 2y. Express y in terms of x, substitute, and minimize P(x).

Example #5: Maximizing Area with Fixed Perimeter (Rectangle)

Problem: Find the dimensions of a rectangle with a perimeter of 76 feet that has the maximum area.

• Solution: Dimensions are 19 ft by 19 ft.

Explanation: For perimeter P = 2x + 2y = 76, area A = xy. Express y in terms of x, substitute, and maximize A(x).

Example #6: Maximizing Volume of a Box Formed from a Square

Problem: A box is to be made from a 12 by 12 inch piece of cardboard by cutting equal squares from each corner and folding up the sides. What size squares should be cut to maximize the volume?

• Solution: Squares of 2 inches should be cut from each corner.

Explanation: Let x be the side of the square cut from each corner. The volume is $V(x) = (12 - 2x)^2 x$. Maximize V(x) with respect to x.

Summary Table: Optimization Problem Types

Problem Type	Quantity to Optimize	Constraint	Typical Solution Method
Revenue Maximization	Revenue $R(x)$	Price-demand function	Express R(x), find critical points
Profit Maximization	Profit $P(x)$	Cost and price-demand functions	Express P(x), find critical points
Area/Perimeter Problems	Area or Perimeter	Fixed perimeter or area	Express in one variable, optimize
Box Volume Problems	Volume $V(x)$	Fixed material size	Express V(x), find maximum