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.

The dimensions of the field that will maximize the enclosed area are 350/9 ft and 17.5 ft and this can be determined by forming the linear equation.

Given :

Let 'x' be vertical, and 'y' be horizontal. So, the linear equation becomes:

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

Simplify the above expression.

700 = 20y + 9x

Now, solve the above equation for 'y'.

   --- (1)

Now, the formula of the area of the rectangle is:

A = xy

Now, substitute the value of 'y' in the above formula.

Now, differentiate the above equation with respect to 'x' and then equate to 0.

Now, equate the above equation to zero.

35 - 0.9x = 0

x = 350/9

Now, substitute the value of 'x' in equation (1).

y = 17.5 ft

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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

a = 11 : 4

b = 38

c = 1 for every 6