Application - Optimization

Question

Suppose that 231 ft of fencing are used to enclose a corral in the shape of a rectangle with a semicircle whose diameter is a side of the rectangle as the following figure:

Find the dimensions of the corral with maximum area.

x=  ft.

y=  ft.

Answer 1

Hello,

Let's first describe the perimeter, P, of this fenced area (which is equal to the length of fencing used). Here I am just simply adding the rectangular sides and remembering that the circumference of a full circle is πx. Then I am also remembering that we have a half circle here and not a full circle.

P = x + 2y + (π x)/2 = 231

(we are told that they are using 231 feet of fencing)

Let's also describe the area, A, enclosed by this fencing. Here I am just adding the area of the rectangular part to the area of the half-circle part.

A = xy + (π x²)/8

Now we need to maximize this area by taking the derivative with respect to one of the variables and setting that derivative equal to zero. But, there's a problem here because our equation A is expressed both in terms of x and y. Let's make it based on x only by relying on the Perimeter description we first wrote down.

P = x + 2y + (π x)/2 = 231

We can solve this for y to get

y = (231/2) - (x/2)(1 + π/2)

Now, we plug this y into the area equation

A = x ((231/2) - (x/2)(1 + π/2)) + (π x²)/8

(we now have area described only in terms of x)

Now differentiate this with respect to x, and set this derivative equal to zero. We are doing this because at this point (where the derivative is zero), the area function will be at it's peak value.

dA/dx = (231/2) - (1 + π/2)x + (2π/8)x = 0

Now just solve this for x. This x is one of your answers (the value of x that maximizes the area). To get y, simply just plug in this solved x value into the description we made: y = (231/2) - (x/2)(1 + π/2)