Question

You have $6000 with which to build a rectangular enclosure with fencing. The fencing material costs $30 per meter. You also want to have two partitions across the width of the enclosure, so that there will be three separated spaces in the enclosure. The material for the partitions costs $25 per meter. What is the maximum area you can achieve for the enclosure? (Round your answer to the nearest whole number.)

Answer 1

Refer to the figure shown below.
x =  the width of the rectangle (meters)
y = the height of the rectangle (meters(

The fencing for the perimeter of the rectangle costs  $30 per meter.
The two inner partitions cost $25 per meter.

The total cost of the fencing is
C = 2(x+y)*$30 + 2y*$25
   = 60(x+y) + 50y 
   = 60x + 110y

Because the amount available to spend is $600, therefore
60x + 110y = 6000
or
6x + 11y = 600
That is,
y = (600 - 6x)/11              (1)

The area is
A = x*y                           (2)

Substitute (1) into (2).
A = (x/11)*(600 - 6x) = (1/11)*(600x - 6x²)

To maximize A, the derivative of A with respect to x is zero.
That is,
600 - 12x = 0
x = 600/12 = 50
From (1), obtain
y = (1/11)*(600 - 6*50) = 300/11 = 27.273

Because the second derivative of A with respect to x is negative, x=50, y = 27.273 will yield the maximum area.

The maximum area is
50*27.273 = 1363.64 m² = 1364 m² (nearest integer)

Answer: 1364 m² (nearest integer)