Answer:
[tex]Length = 500ft[/tex]
[tex]Width = 250ft[/tex]
Step-by-step explanation:
Given
[tex]P = 1000[/tex] --- perimeter
Required
The dimensions that maximize the area
Let
[tex]L \to Length[/tex]
[tex]W \to Width[/tex]
Such that:
[tex]P = L + 2W[/tex] ---- Because one side is from the school building
This gives:
[tex]L + 2W = 1000[/tex]
Make L the subject
[tex]L = 1000 - 2W[/tex]
The area is then calculated as:
[tex]A= L * W[/tex]
Substitute: [tex]L = 1000 - 2W[/tex]
[tex]A= (1000 - 2W) * W[/tex]
Open bracket, then set A to 0
[tex]A = 1000W - 2W^2[/tex]
[tex]1000W- 2W^2 = 0[/tex]
Rewrite as:
[tex]-2W^2 + 1000W = 0[/tex]
Assume the form of the above equation is: [tex]aW^2 + bW + c = 0[/tex]
The value of W that maximizes it is:
[tex]W = -\frac{b}{2a}[/tex]
Where
[tex]a = -2[/tex]
[tex]b = 1000[/tex]
[tex]c = 0[/tex]
So:
[tex]W = -\frac{1000}{2*-2}[/tex]
[tex]W = \frac{1000}{2*2}[/tex]
[tex]W = \frac{1000}{4}[/tex]
[tex]W = 250[/tex]
Recall that:
[tex]L = 1000 - 2W[/tex]
[tex]L =1000 - 2 * 250[/tex]
[tex]L =1000 - 500[/tex]
[tex]L = 500[/tex]
So, the dimensions that maximize the area is:
[tex]Length = 500ft[/tex]
[tex]Width = 250ft[/tex]