### Introduction to Optimization Problems: AP Calculus AB/BC Study Guide

#### What’s the Deal with Optimization?

Welcome to the wild world of optimization problems in AP Calculus, where we use fancy math tricks to find the best possible solution—think of it as the calculus version of finding the perfect pizza slice! 🍕 Optimization problems aim to maximize or minimize a particular quantity, whether it's maximizing profits or minimizing costs, or even determining the optimal amount of sleep before an exam (hint: it’s never zero).

#### Maximizing, Minimizing, and Derivatives, Oh My!

Previously, we learned to identify the minimum or maximum values of functions using the First or Second Derivative Tests. These tools are like the Swiss Army knives of calculus, helping us pinpoint those critical values where something interesting happens—like reaching a peak or hitting rock bottom.

Here’s the key to cracking optimization problems: Understand that these problems often come with extra variables, like unplanned plot twists in a TV show. Our job? Simplify things by finding a relationship between these variables, which lets us boot out the extras and focus on just one—voilà, optimization becomes manageable! 🧙♂️✨

#### A Case of Optimization: Walkthrough

Let's dive into an example to see how this magic unfolds:

#### Problem: Minimizing a Sum of Squares

Consider ( s = x^2 + y^2 ) where ( xy = -36 ). We need to find values of ( x ) and ( y ) that minimize ( s ).

**Simplify the Problem:**

Since we have two variables ( x ) and ( y ), but differentiation works best with one, we'll express ( y ) in terms of ( x ). Given ( xy = -36 ):

[ y = -\frac{36}{x} ]

Now substitute ( y ) into ( s ):

[ s(x) = x^2 + \left(-\frac{36}{x}\right)^2 = x^2 + \frac{1296}{x^2} ]

**Find the Critical Points:**

We first need the derivative ( s'(x) ):

[ s'(x) = 2x - \frac{2592}{x^3} ]

Set ( s'(x) = 0 ) to find critical points:

[ 2x - \frac{2592}{x^3} = 0 ]

Multiply through by ( x^3 ):

[ 2x^4 = 2592 ] [ x^4 = 1296 ] [ x = \pm 6 ]

**Determine Concavity:**

We need the second derivative ( s''(x) ):

[ s''(x) = 2 + \frac{7776}{x^4} ]

Evaluate ( s''(x) ) at ( x = \pm 6 ):

[ s''(6) = 2 + \frac{7776}{6^4} ]

Since ( s''(x) > 0 ) for all ( x \neq 0 ), ( s(x) ) is concave up at ( x = \pm 6 ). Therefore, ( x = \pm 6 ) are minimum points.

**Find Corresponding ( y ) Values:**

For ( x = 6 ):

[ y = -\frac{36}{6} = -6 ]

For ( x = -6 ):

[ y = -\frac{36}{-6} = 6 ]

Thus, the solutions are ((6, -6)) and ((-6, 6)). These pairs minimize ( s(x) ).

#### More Practice, Because Practice Makes Perfect

Let's do some additional problem-solving to reinforce this.

##### Problem 1: Play Area Optimization

An open-topped play area with a square base must hold 32 cubic feet of sand. Find the minimum exterior surface area.

Here's how we solve it:

**Volume Constraint:**

[ V = x^2h = 32 ] [ h = \frac{32}{x^2} ]

**Surface Area Expression:**

[ S(x) = x^2 + 4x \cdot \frac{32}{x^2} = x^2 + \frac{128}{x} ]

**Find Critical Points:**

[ S'(x) = 2x - \frac{128}{x^2} ] [ 2x^3 = 128 ] [ x^3 = 64 ] [ x = 4 ]

**Check Concavity:**

[ S''(x) = 2 + \frac{256}{x^3} ] [ S''(4) > 0 ]

Thus, ( x = 4 ) minimizes ( S(x) ), giving a minimum surface area of 48 square feet.

##### Problem 2: Printing Problem

Print 121 square inches of text with one-inch margins on all sides. Find the dimensions of the smallest card.

**Area Constraint:**

[ xy = 121 ] [ y = \frac{121}{x} ]

**Card Dimensions:**

[ \text{Width} = x + 2 ] [ \text{Height} = \frac{121}{x} + 2 ]

**Area Expression:**

[ A(x) = (x + 2) \left(\frac{121}{x} + 2\right) ]

**Find Critical Points:**

[ A'(x) = 2 - \frac{242}{x^2} ] [ x^2 = 121 ] [ x = 11 ]

**Check Concavity:**

[ A''(x) = \frac{484}{x^3} > 0 ]

Thus, the smallest card dimensions are 13 inches by 13 inches.

#### Recap and Key Terms

Great job sticking with it! 🎉 You've tackled some tough optimization problems and come out on top. Here are a few key terms to remember:

**Minimum**: The lowest value a function or variable can attain.**Optimization**: The process of finding the best possible solution.**Relationship**: The connection between two or more variables.

#### Final Thoughts

Fantastic work so far! You're almost finished with Unit 5 of AP Calculus. Keep practicing to master optimization problems like a pro. Next up, we'll dive into more examples to sharpen your skills. 📚✨

Happy optimizing! 🚀