1. 4.4 Modeling and Optimization Buffalo Bill’s Ranch, North Platte, Nebraska Greg Kelly, Hanford High School, Richland, WashingtonPhoto by Vickie Kelly, 1999

2. Optimization Applying Calculus and its methods for finding minimums and maximums in real life situations such as What is the largest area I can enclose with a given amount of fence? What is the least amount of material needed to build a cylinder of given volume? What dimensions will maximize the volume of a rectangular prism with given surface area?

3. Basic Steps to Solve Optimization Problems 1.Read the problem 2.Draw a picture to help you visualize the situation 3.Write 1 equation for each variable. (You should have the same number of equations as variables)

4. 4. Take the equation with the number so you can solve for one of the variables 5. Substitute the variable you found into one of the other equations 6. Take the derivative and set the equation equal to 0 Basic Steps to Solve Optimization Problems

5. 7. Take the second derivative to check your answer (if it’s 0 min) 8. Solve for the other variables needed to answer the question

6. A Classic Problem You have 40 feet of fence to enclose a rectangular garden along the side of a barn. What is the maximum area that you can enclose? There must be a local maximum here, since the endpoints are minimums. (and the function is a downward parabola)

7. A Classic Problem You have 40 feet of fence to enclose a rectangular garden along the side of a barn. What is the maximum area that you can enclose?

8. General procedures for finding solutions to optimization problems: 1Draw a picture and appropriately label the important parts. 2.Write an equation (function) for the quantity you want to optimize in terms of one variable. 3 Find and test all critical numbers. (answers are usually fairly obvious.)

9. What is the largest open top box I can make by cutting equal squares from each corner of a 6x6 sheet of metal & turning up the sides? V(x) = x(6 - 2x)(6 - 2x) = 36x – 24x 2 + 4x 3 V’(x) = 36 – 48x + 12x 2 12x 2 – 48x + 36 = 0 x 2 – 4x + 3 = 0 (x – 3)(x – 1) = 0 x = 3 or x = 1 V(1) = 16 cubic units V(x) = x(6 - 2x)(6 - 2x) V(1) = (1)(6 - 2)(6 - 2)

10. Example 5: What dimensions for a one liter cylindrical can will use the least amount of material? We can minimize the material by minimizing the surface area. area of ends lateral area We need another equation that relates r and h :

11. Example 5: What dimensions for a one liter cylindrical can will use the least amount of material? area of ends lateral area

12. If the end points could be the maximum or minimum, you have to check. (This is rare) To summarize: Write a function for the amount that you want to optimize. If you are not sure that the extreme you’ve found is a maximum or a minimum, you have to check.  If the function that you want to optimize has more than one variable, use substitution to rewrite the function.