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diamong [38]
3 years ago
13

An engineer must design a rectangular box that has a volume of 9 m3 and that has a bottom whose length is twice its width. What

are the dimensions of the box so that the total surface area (of all six sides) of the box is minimized
Engineering
1 answer:
Jet001 [13]3 years ago
6 0

Answer:

Length =3   Height = 2   and  Width = \frac{3}{2}

Explanation:

Given

Volume = 9m^3

Represent the height as h, the length as l and the width as w.

From the question:

Length = 2 * Width

l = 2w

Volume of a box is calculated as:

V = l*w*h

This gives:

V = 2w *w*h

V = 2w^2h

Substitute 9 for V

9 = 2w^2h

Make h the subject:

h = \frac{9}{2w^2}

The surface area is calculated as:

A = 2(lw + lh + hw)

Recall that: l = 2w

A = 2(2w*w + 2w*h + hw)

A = 2(2w^2 + 2wh + hw)

A = 2(2w^2 + 3wh)

A = 4w^2 + 6wh

Recall that: h = \frac{9}{2w^2}

So:

A = 4w^2 + 6w * \frac{9}{2w^2}

A = 4w^2 + 6* \frac{9}{2w}

A = 4w^2 + \frac{6* 9}{2w}

A = 4w^2 + \frac{3* 9}{w}

A = 4w^2 + \frac{27}{w}

To minimize the surface area, we have to differentiate with respect to w

A' = 8w - 27w^{-2}

Set A' to 0

0 = 8w - 27w^{-2}

Add 27w^{-2} to both sides

27w^{-2} = 8w

Multiply both sides by w^2

27w^{-2}*w^2 = 8w*w^2

27 = 8w^3

Make w^3 the subject

w^3 = \frac{27}{8}

Solve for w

w = \sqrt[3]{\frac{27}{8}}

w = \frac{3}{2}

Recall that : h = \frac{9}{2w^2}   and l = 2w

h = \frac{9}{2 * \frac{3}{2}^2}

h = \frac{9}{2 * \frac{9}{4}}

h = \frac{9}{\frac{9}{2}}

h = 9/\frac{9}{2}

h = 9*\frac{2}{9}

h= 2

l = 2w

l = 2 * \frac{3}{2}

l = 3

Hence, the dimension that minimizes the surface area is:

Length =3   Height = 2   and  Width = \frac{3}{2}

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