Answer:
[tex]l=10[/tex] [tex]w=20[/tex]
Step-by-step explanation:
Given
[tex]l \to length[/tex]
[tex]w \to width[/tex]
[tex]A= 200ft^2[/tex]
Required
The dimension to minimize cost
Area is calculated as:
[tex]A =l * w[/tex]
This gives:
[tex]l * w = 200[/tex]
Make l the subject
[tex]l = \frac{200}{w}[/tex]
The perimeter P is:
[tex]P = 2l + w[/tex] ---- because one part will not be covered
Substitute: [tex]l = \frac{200}{w}[/tex]
[tex]P = 2 * \frac{200}{w} + w[/tex]
[tex]P = \frac{400}{w} + w[/tex]
Rewrite as:
[tex]P = 400w^{-1} + w[/tex]
Differentiate
[tex]P' = -400w^{-2} + 1[/tex]
Set to 0
[tex]-400w^{-2} + 1 = 0[/tex]
Rewrite as:
[tex]-400w^{-2} =- 1[/tex]
Divide by -1
[tex]400w^{-2} =1[/tex]
Rewrite as:
[tex]\frac{400}{w^2} =1[/tex]
Solve for [tex]w^2[/tex]
[tex]w^2 = 400[/tex]
Take square roots
[tex]w=20[/tex]
Recall that: [tex]l = \frac{200}{w}[/tex]
[tex]l =200/20[/tex]
[tex]l=10[/tex]
