A rectangular building covering a 7000 m2 floor space is to

A rectangular building covering a 7000 m^2 floor space is to be constructed on a rectangular piece of land as shown in the diagram. If the building is to be 10.0 m from the boundary of the land on each side, and 20.0 m from the boundary at the front and back, what should be the dimensions of the building if the area of the lot is to be minimized? Ensure that you define all variables that you introduce, explain your working and verify that the dimensions you find actually provide a minimum lot area.

Solution

Let the length of the building be x and the width be y.

Then, the area of the building = xy = 7000 (given)....... (1)

Given xy = 7000, y = 7000/x ......... (2)

Now, the dimensions of the lot will be: length = (x + 20 + 20)

= x + 40

and width =: (y + 10 + 10)

= y + 20.

Hence the area of the lot, say A = (x + 40)(y + 20)

= xy + 20x + 40y + 800

= 7000 + 20x + 40y + 800[substituting from equation (1) above]

= 20x + 40y + 7800

= 20x + 40(7000/x)[substituting from equation (2) above] + 7800

= 20x + (28000/x) + 7800.

Thus, we have, A = 20x + (28000/x) + 7800.

Since it is required to minimise the lot area, the first derivative of A must be zero and second derivative must be positive. This implies: 20 - (28000/x^2) = 0 .... (3) and

(2 x 28000)/x1^3 > 0, where x1 is the value of x obtained from equation (3) above.

From equation (3), x^2 = 14000 or x ~ 118.5 i.e., x1 = 118.5 ..... (4).

Since x1 is positive, (2 x 28000)/x1^3 > 0.

Thus, the lot area minimising length ~ 118.5 and the corresponding width, y, is given by:

y^2 = 7000^2/14000

= 3500 or

y ~ 59.5.

Answer: The area optimising dimensions of the building are: length = 118.5m(approx) and width = 58.5(approx).