Optimization

Optimization

 * Optimization problems include:
 * Recognizing equations which relate the information given to the information that is desired
 * Using derivatives and sign charts to locate maximums and minimums
 * Algebraically solving for variables.

**Strategy for Solving Max-Min Problems**

 * **Understand the Problem**
 * Read the Problem carefully. Identify the information you need to solve the problem.
 * ** Develop a Mathematical Model of the Problem **
 * ** ﻿ ** Draw pictures and label the parts that are important to the problem. Introduce a variable to represent the quantity to be maximized or minimized. Using that variable, writea function whose extreme value gives the information sought.
 * ** Graph the Function **
 * Find the domain of the function. Determine what values of the variable make sense in the problem.
 * ** Identify the Critical Points and Endpoints **
 * Find where the derivative is zero or fails to exist.
 * ** Solve the Mathematical Model **
 * If unsure of the result, support or confirm your solution with another method.
 * ** Interpret the Solution **
 * Translate your mathematical result into the problem setting and decide whether the result makes sense.

Example Problem
A rectangular plot of famland will be bounded on one side by a river and on the other three sides by a single strand electric fence. With 800 m of wire at your disposal, what is the largest area you can enclose, and what are its dimensions?


 * Decide what is being optimized and write an equation:
 * In this problem area is the target for optimization. The general equation for area is A=lw.
 * There are two variables in this problem ( Length and Width) so two equations are needed.
 * A value is given for perimeter so the equation P=2l+2w can be used to relate area and perimeter. Because one of the barriers is going to be formed by the river one of the coefficients can be changed to 1. P=2l +w.
 * Substitute the value for perimeter into the equation and solve for a variable.
 * (800)=2l+w
 * 800-2l=w
 * Substitute the solved variable into the original area equation.
 * A=lw
 * A=l(800-2l)
 * Take the derivative of the new equation:
 * A'=l(-2)+(800-2l)
 * Set the derivative equal to zero and solve for the single variable.
 * l(-2)+(800-2l)=0
 * -2l +800-2l=0
 * -4l+800=0
 * -4l=-800
 * l=200
 * Construct a sign chart to find where a maximum will occur.
 * [[file:sin chart.docx]]


 * Maximum will occur when l=200.
 * Substitute 200 into the perimeter equation to find the value of w when l=200
 * P=2l+w
 * 800=2(200)+w
 * 800=400+w
 * 400=w
 * Substitute values for both variables into the Area equation to find the max area.
 * A=lw
 * A=(200)(400)
 * A=80,000

Works Cited Finney, R.L., Demana, F.D., Waits, B.K., & Kennedy, D. (2007). //Calculus: graphical, numerical, algebraic//. Boston, Massachusetts: Pearson Prentice Hall.