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A closed box has a square base with side length L feet and height h feet. Given that the volume of the box is 39 cubic feet, express the surface area of the box in terms of L only. Need help please?

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Explanation:

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May 19, 2017

$= \left(2 {L}^{2} + \frac{156}{L}\right) f {t}^{2}$

Explanation:

Given that the volume of the box $V = 39 f {t}^{3}$

Again $\text{volume "V= "area of base"xx"height} = {L}^{2} \times h$

So $h = \frac{V}{L} ^ 2 = \frac{39}{L} ^ 2$

Now surface area of the box
$S = 2 \left({L}^{2} + L \times h + L \times h\right) = 2 \left({L}^{2} + 2 \times L \times h\right)$

$= 2 \left({L}^{2} + 2 \times L \times \frac{39}{L} ^ 2\right)$

$= \left(2 {L}^{2} + \frac{156}{L}\right) f {t}^{2}$

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