Answer:
[tex]Length = Width = 6\ units[/tex]
Step-by-step explanation:
Given
[tex]Area=36[/tex]
Required
The least possible material
Sandboxes usually, are rectangles or squares.
Using the above assumption, the area is calculated as:
[tex]Area= Length * Width[/tex]
[tex]Area= L* W[/tex]
[tex]L * W = 36[/tex]
Make L the subject
[tex]L = \frac{36}{W}[/tex]
The material of the outer edge is calculated by the perimeter.
[tex]Perimeter = 2 * (L +W)[/tex]
[tex]P = 2 * (L + W)[/tex]
Substitute [tex]L = \frac{36}{W}[/tex]
[tex]P = 2 * (\frac{36}{W} + W)[/tex]
Open bracket
[tex]P = \frac{72}{W} + 2W[/tex]
[tex]P = 72W^{-1} +2W[/tex]
To get the minimum material needed, we differentiate P
[tex]P' = -72W^{-2} + 2[/tex]
Set: [tex]P' = 0[/tex]
[tex]-72W^{-2} + 2 = 0[/tex]
Collect like terms
[tex]72W^{-2} = 2[/tex]
Divide both sides by 72
[tex]W^{-2} = \frac{2}{72}[/tex]
[tex]W^{-2} = \frac{1}{36}[/tex]
Rewrite as:
[tex]\frac{1}{W^2} = \frac{1}{36}[/tex]
Take positive square roots of both sides
[tex]\frac{1}{W} = \frac{1}{6}[/tex]
Cross multiply
[tex]W = 6[/tex]
Recall that: [tex]L = \frac{36}{W}[/tex]
So, we have:
[tex]L =\frac{36}{6}[/tex]
[tex]L = 6[/tex]
Hence, the dimension with the littlest material as possible is:
[tex]Length = Width = 6\ units[/tex]
