Question

A rancher wishes to build a fence to enclose

a 800 square yard rectangular field. Along one
side the fence is to be made of heavy duty ma-
terial costing $9 per yard, while the remaining
three sides are to be made of cheaper mate-
rial costing $3 per yard. Determine the least
cost, Cmin, of fencing for the rancher.

Answer 1

Explanation:

Let's call the length of the field "L" and the width of the field "W".

We know that the area of the rectangular field is 800 square yards, so:

L x W = 800

We want to minimize the cost of the fencing, which is made up of two different types of material. Let's call the length of the side with the expensive material "x".

The cost of the expensive side is 9x, and the cost of the other three sides is 3(2L + 2W - x). This is because there are two lengths and two widths, but we're subtracting out the length of the expensive side (x) since we're already accounting for it separately.

So the total cost of the fencing is:

C = 9x + 3(2L + 2W - x)

Now we can use the area equation (L x W = 800) to solve for one of the variables. Let's solve for W:

W = 800/L

Now we can substitute this into our cost equation:

C = 9x + 3(2L + 2(800/L) - x)

Simplifying this equation:

C = 9x + 6L + 4800/L - 3x

C = 6L + 6x + 4800/L

To minimize this function, we can take the derivative with respect to L and set it equal to zero:

dC/dL = 6 - 4800/L^2 = 0

Solving for L:

L = 40

Now we can use the area equation to solve for W:

W = 800/40 = 20

Finally, we can use these values to find the length of the expensive side:

x = L = 40

So the minimum cost of the fencing is:

Cmin = 9x + 3(2L + 2W - x)

Cmin = 9(40) + 3(2(40) + 2(20) - 40)

Cmin = $660