# area of a box, optimization problem.

I recognize that this is an optimization problem and I need to take the derivative set it equal to 0 and plug it back into the original function but I am having trouble figuring out the original function.

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Let the height, width and length of the box be $h,w,l$ respectively. Since the top and bottom are squares, let $l = w = a$. Therefore, the volume of the box is $ha^2 = 14.71$ and the total material cost is $f(a,h) = 2a^2 + 4ah$ which is the area of the top and bottom ($2a^2$) and the 4 sides ($4ah$).
Hint: You can reduce $f$ to a function of a single variable ($g(a)$ or $g(h)$) using $ha^2 = 14.71$ and then find its derivative $g'$.