A box with a square base and open top must have a volume of 318028 $\displaystyle {c}{m}^{{3}}$. We wish to find the dimensions of the box that minimize the amount of material used.

First, find a formula for the surface area of the box in terms of only $\displaystyle {x}$, the length of one side of the square base.
[Hint: use the volume formula to express the height of the box in terms of $\displaystyle {x}$.]
Simplify your formula as much as possible.
$\displaystyle {A}{\left({x}\right)}=$Preview Question 6 Part 1 of 5  

Next, find the derivative, $\displaystyle {A}'{\left({x}\right)}$.
$\displaystyle {A}'{\left({x}\right)}=$Preview Question 6 Part 2 of 5  

Now, calculate when the derivative equals zero, that is, when $\displaystyle {A}'{\left({x}\right)}={0}$. [Hint: multiply both sides by $\displaystyle {x}^{{2}}$.]
$\displaystyle {A}'{\left({x}\right)}={0}$ when $\displaystyle {x}=$

We next have to make sure that this value of $\displaystyle {x}$ gives a minimum value for the surface area. Let's use the second derivative test. Find A"$\displaystyle {\left({x}\right)}$.
A"$\displaystyle {\left({x}\right)}=$Preview Question 6 Part 4 of 5  

Evaluate A"$\displaystyle {\left({x}\right)}$ at the $\displaystyle {x}$-value you gave above.

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NOTE: Since your last answer is positive, this means that the graph of $\displaystyle {A}{\left({x}\right)}$ is concave up around that value, so the zero of $\displaystyle {A}'{\left({x}\right)}$ must indicate a local minimum for $\displaystyle {A}{\left({x}\right)}$. (Your boss is happy now.)