Michael J. answered 06/22/17
Tutor
5
(5)
Mastery of Limits, Derivatives, and Integration Techniques
Let width = w
Let length = x
You will need to use these equations in the optimization:
2(x + w) = 264
x + w = 132 eq1
Area = xw eq2
You want to find the maximum area, so substitute the perimeter equation (eq1) into the area equation (eq2). This will get the area equation in terms of only x.
Area = x(132 - x)
Area = 132x - x2
Then, take the derivative of Area and set it equal to zero.
d/dx (Area) = 0
132 - 2x = 0
Solve for x to get the critical value. This critical value is the length that will give the maximum area. Then once you have the length, plug it into eq1 to get the width.