Quadratic equation

Question

A piece of land that already fenced up by using a wire fence of 120 m. Both of the rectangular lands are placed next to each other with both lands have equal length, x m. and width, y m, respectively.

(a) Determine the values of x and y such that the whole area of the land is maximum.

(b) Find the maximum area of the land.

Answer 1

Perimeter of the land is 2(x+y) = 120

so x+y = 60

The area of two equal pieces of land are 2*(x^2y^2)

We need to maximize Area = x^y^2 or x^2(60-x)^2

We need to take first derivative of A with respect to x to maximize area

dA/dx = 2x(60-x)^2 + 2x^2(60-x)(-1) = 0

x(60-x)^2 = x^2(60-x)

we can cancel one x on each side as x= 0 will not be a solution

(60-x)^2 = x(60-x)

3600+x^2-120x = 60x -x^2

2x^2 -180x+3600 = 0

x^2-90x+1800 = 0

x^2-30x-60x+1800 = 0

or the solution is x= 30 or x = 60

Since x+y=60

Y can not be 0 and hence x can not be 60, and hence x= 30 and y = 30 (final solution)

so the maximum area is when both x and y are equal and hence both pieces are squares.