## Calculus Tutorial

*First Derivative Max/Min Optimization *

#### Intro

This is one of the most popular concepts that you will be asked in your college calculus courses. Finding maximum and minimum values of a given equation or graph. This type of problem often requires you to craft an equation with only one variable for taking the first derivative so you can find max or min of the equation in order to solve it. On the other hand, you might be given a graph where you’ll need to find the equation by plugging in x and y values in your Ti 83/89 calculator.

#### Sample Problem

Find two non-negative numbers whose sum is 15 and so that the product of one number and the square of the other number is a maximum.

#### Solution

1. Let variables x and y represent two nonnegative numbers. The sum of the two numbers is given to be 15 = x + y, so y = 15 – x.

2. We want to find the maximum of the product of two numbers. So P = x y^2. However, we need to condense the equation into one variable only as stated in the intro. From step 1, we know y = 15 – x, so P = x ( 15-x)^2

3. Using both chain and product rules in your calculus textbook, differentiate the equation in step 2:

P’ = x (2) ( 15-x)(-1) + (1) ( 15-x)^2

= ( 15-x) [ -2x + (15-x) ]

= ( 15-x) [ 15-3x ]

= ( 15-x) (3)[ 5-x ]

= 0

So x = 15 or x = 5.

Note that since both x and y are non-negative numbers and their sum is 15, it follows that 0 <= x <= 15
4. y = 15 -x = 15 - 5 = 10
5. Finally, find the product from step 2: P = x ( 15-x)^2 = 5 * 10^(2) = 500(answer).

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