now, check the 1st picture below. we know the original paperboard is a 16x14, so the width, as you see in the picture is w = 8 - x, its length is L = x, and its height is 14 -(x/2) - (x/2).

$\bf V(x)=\stackrel{L}{(x)}\stackrel{w}{(8-x)}\stackrel{h}{\left( 14-\cfrac{x}{2}-\cfrac{x}{2} \right)}\implies V(x)=(8x-x^2)(14-x) \\\\\\ V(x)=112x-8x^2-14x^2+x^3\implies V(x)=x^3-22x^2+112x$

now, let's find the critical points, by setting the derivative to 0, and then we'll do a first-derivative test.

$\bf \cfrac{dV}{dx}=3x^2-44x+112\implies 0=3x^2-44x+112 \\\\\\ \textit{plugging those values in the quadratic formula}\qquad x\approx \begin{cases} 11.39\\ 3.28 \end{cases}$

now, if we plot those two points, we get 3 regions, and then we check each of those regions to see what's the value of the first derivative.

check the 2nd picture below, that'd be the first derivative test, as you can see from the arrows, the slope increases and then decreases at 3.28, namely that critical point is the maximum, so the Volume maximizes at x = 3.28.

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2 years ago