Answer:
[tex]Length =3[/tex] [tex]Height = 2[/tex] and [tex]Width = \frac{3}{2}[/tex]
Explanation:
Given
[tex]Volume = 9m^3[/tex]
Represent the height as h, the length as l and the width as w.
From the question:
[tex]Length = 2 * Width[/tex]
[tex]l = 2w[/tex]
Volume of a box is calculated as:
[tex]V = l*w*h[/tex]
This gives:
[tex]V = 2w *w*h[/tex]
[tex]V = 2w^2h[/tex]
Substitute 9 for V
[tex]9 = 2w^2h[/tex]
Make h the subject:
[tex]h = \frac{9}{2w^2}[/tex]
The surface area is calculated as:
[tex]A = 2(lw + lh + hw)[/tex]
Recall that: [tex]l = 2w[/tex]
[tex]A = 2(2w*w + 2w*h + hw)[/tex]
[tex]A = 2(2w^2 + 2wh + hw)[/tex]
[tex]A = 2(2w^2 + 3wh)[/tex]
[tex]A = 4w^2 + 6wh[/tex]
Recall that: [tex]h = \frac{9}{2w^2}[/tex]
So:
[tex]A = 4w^2 + 6w * \frac{9}{2w^2}[/tex]
[tex]A = 4w^2 + 6* \frac{9}{2w}[/tex]
[tex]A = 4w^2 + \frac{6* 9}{2w}[/tex]
[tex]A = 4w^2 + \frac{3* 9}{w}[/tex]
[tex]A = 4w^2 + \frac{27}{w}[/tex]
To minimize the surface area, we have to differentiate with respect to w
[tex]A' = 8w - 27w^{-2}[/tex]
Set A' to 0
[tex]0 = 8w - 27w^{-2}[/tex]
Add [tex]27w^{-2}[/tex] to both sides
[tex]27w^{-2} = 8w[/tex]
Multiply both sides by [tex]w^2[/tex]
[tex]27w^{-2}*w^2 = 8w*w^2[/tex]
[tex]27 = 8w^3[/tex]
Make [tex]w^3[/tex] the subject
[tex]w^3 = \frac{27}{8}[/tex]
Solve for w
[tex]w = \sqrt[3]{\frac{27}{8}}[/tex]
[tex]w = \frac{3}{2}[/tex]
Recall that : [tex]h = \frac{9}{2w^2}[/tex] and [tex]l = 2w[/tex]
[tex]h = \frac{9}{2 * \frac{3}{2}^2}[/tex]
[tex]h = \frac{9}{2 * \frac{9}{4}}[/tex]
[tex]h = \frac{9}{\frac{9}{2}}[/tex]
[tex]h = 9/\frac{9}{2}[/tex]
[tex]h = 9*\frac{2}{9}[/tex]
[tex]h= 2[/tex]
[tex]l = 2w[/tex]
[tex]l = 2 * \frac{3}{2}[/tex]
[tex]l = 3[/tex]
Hence, the dimension that minimizes the surface area is:
[tex]Length =3[/tex] [tex]Height = 2[/tex] and [tex]Width = \frac{3}{2}[/tex]
