Question 990661
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Let *[tex \Large s] represent the measure of the side of the square bottom of the box.  Let *[tex \Large x] represent the height of the box.


The volume of the box is *[tex \Large s^2x\ =\ 144], so solving for *[tex \Large s], we get *[tex \Large s\ =\ \frac{12}{sqrt{x}}]


The surface area of an open box is the area of the 4 sides, *[tex \Large 4sx\ =\ \frac{48x}{\sqrt{x}}\ =\ 48\sqrt{x}] plus the area of the bottom, *[tex \Large s^2\ =\ \frac{144}{x}]


So, the surface area as a function of the height of the box is:


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ S(x)\ =\ 48\sqrt{x}\ +\ \frac{144}{x}]


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ \frac{dS}{dx}\ =\ \frac{24}{sqrt{x}}\ -\ \frac{144}{x^2}]


Set the derivative equal to zero and solve for *[tex \Large x], then use *[tex \Large s\ =\ \frac{12}{sqrt{x}}]. 


John
*[tex \LARGE e^{i\pi}\ +\ 1\ =\ 0]
My calculator said it, I believe it, that settles it

*[tex \Large \ \
*[tex \LARGE \ \ \ \ \ \ \ \ \ \