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4.7 Modeling and Optimization
Buffalo Bill’s Ranch, North Platte, Nebraska Greg Kelly, Hanford High School, Richland, Washington Photo by Vickie Kelly, 1999
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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.
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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.
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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)
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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.
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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.
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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?
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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
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Example 5: What dimensions for a one liter cylindrical can will use the least amount of material? area of ends lateral area
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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.
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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
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