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)
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)