A box with a square base and open top must have a volume of $13, 500 c m^{3}$ .
Find the dimensions of the box that minimize the amount of material used.
sides of base =_cm
height =__cm

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Solution

The Volume of a box with a square base $x b y x c m$ and height $h c m$ is $V = x^{2} h$
The amount of material used is directly proportional to the surface area, so we will minimize the amount of material by minimizing the surface area.
The surface area of the box described is $A = x^{2} + 4 x h$
We need A as a function of x alone, so we'll use the fact that
$V = x^{2} h = 13, 500 c m^{3}$
which gives us $h = (\frac{13500}{x^{2}})$ so the area becomes:
$A = x^{2} + 4 x (\frac{13500}{x^{2}}) = x^{2} + \frac{54000}{x}$
We want to minimize $A$ , so
$A' = 2 x - \frac{54, 000}{x^{2}} = 0 w h e n \frac{2 x^{3} - 54000}{x^{2}} = 0$
Which occurs when $x^{3} - 27, 000 = 0 o r x = 30$
The only critical number is $x = 30 c m .$
The second derivative test verifies that A has a minimum at this critical number:
$A'' = 2 + \frac{108000}{x^{3}} w h i c h i s p o s i t i v e a t x = 30 .$
The box should have base $30 c m b y 30 c m$ and height $15 c m$
Hence, the base should be $30 c m b y 30 c m$ and height should be $15 c m$

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