Question

We are going to fence in a rectangular field. If we look at the field from above the cost of the vertical sides are $10/f t, the cost of the bottom is $2/f t and the cost of the top is $7/f t. If we have 700 determine the dimensions of the field that will maximize the enclosed area.

Answer 1

Answer:

Dimensions are 350/9 ft and 17.5 ft

Step-by-step explanation:

We are given the cost per ft of all the 4 sides. Let the horizontal be x and the vertical be y.

Now, we will set up the constraint and equation that we are being asked to maximize.

Thus;

700 = 10y + 10y + 7x + 2x

700 = 20y + 9x

Maiking y the subject, we have;

y = (700 - 9x)/20

y = 35 - 9x/20

Now,area of a rectangle is: A = xy

Thus, A = x(35 - 9x/20))

A = 35x - 9x²/20

We can get the critical points by finding the derivatives and Equating to zero

Thus;

dA/dx = 35 - 0.9x

At dA/dx = 0,we have; x = 350/9

At d²A/dx², we have;

d²A/dx² = -0.9

This is negative, thus we will disregard and use the one gotten from the first derivative.

Thus, we will use x = 350/9 ft

Plugging this into the equation y = 35 - 9x/20, we have;

y = 35 - ((9 × 350/9)/20)

y = 17.5 ft