4.7 Modeling and Optimization Buffalo Bill’s Ranch, North Platte, Nebraska Greg Kelly, Hanford High School, Richland, Washington Photo by Vickie Kelly, 1999

Optimization Optimization problems in calculus often involve the determination of the “optimal” (meaning, the best) value of a quantity.  Very often, the optimization must be done with certain constraints. Optimal values are often either the maximum or the minimum values of a certain function.

Sample problem: Find the maximum area of a rectangle whose perimeter is 100 meters. Step 1: Determine the function that you need to optimize. In the sample problem, we need to optimize the area A of a rectangle, which is the product of its length L and width W. Our function in this example is A = LW.

Step 2: Identify the constraints to the optimization problem Step 2: Identify the constraints to the optimization problem. In our sample problem, the perimeter of the rectangle must be 100 meters. (This will be useful in the next step). Step 3: Express that function in terms of a single variable upon which it depends, using algebra. For this example, we’re going to express the function in a single variable. “L.” A rectangle’s perimeter is the sum of its sides, that is, 100 = 2L + 2W. Solve for W. W=50-L Substitute 50-L into original area. A=L(50-L)

Step 4: Calculate the derivative of the function with respect to a variable. Step 5: Set the function to zero and compute the corresponding variable’s value. 0=50-2L L=25 Step 6: Use the value from Step 5 to calculate the corresponding optimal value of the function. L=25 W=50-L=50-25=25 A=(25)(25)=625 square meters.

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.

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?

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

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

To find the maximum (or minimum) value of a function: 1 Write it in terms of one variable. 2 Find the first derivative and set it equal to zero. 3 Check the end points if necessary.

Notes: If the function that you want to optimize has more than one variable, use substitution to rewrite the function. If you are not sure that the extreme you’ve found is a maximum or a minimum, you have to check. If the end points could be the maximum or minimum, you have to check. p