Question

We are going to fence in a rectangular field that encloses 75 ft2. Determine the dimensions of the field that will require the least amount of fencing material to be used.

Answer 1

Answer:

For width W and length L, we have:

W = L = √(75ft^2) = 8.66ft

Step-by-step explanation:

For a rectangle of length L and width W, the area is:

A = L*W

and the perimeter is:

P = 2*L + 2*W

First, we know that the area of our rectangle is:

A = 75ft^2 = L*W

And we want to minimize the perimeter of our rectangle, then we need to minimize:

P = 2*L + 2*W

From the equation:

75ft^2 = L*W

We can isolate one of the variables, let's isolate L

L = (75ft^2)/W

We could replace this in the perimeter equation:

P(W) = 2*( (75ft^2)/W) + 2*W

P(W) = (150 ft^2)/W + 2*W

To find the minimum of the perimeter we need to look at the zero of the first derivative of P(W).

P'(W) = dP(W)/dW = -(150ft^2)/W^2 + 2

Now we need to find the value of W such that:

P'(W) = 0

0 =  -(150ft^2)/W^2 + 2

(150 ft^2)/W^2 = 2

(150ft^2) = 2*W^2

(150ft^2)/2 = W^2

75 ft^2 = W^2

√(75ft^2) = W = 8.66ft

And remember that:

L =  (75ft^2)/W

replacing with W = √(75ft^2)

L =  (75ft^2)/√(75ft^2) = √(75ft^2)

Then:

W = L = √(75ft^2) = 8.66ft