A rectangular box with an open top is to be constructed from a 12-foot by 16-foot sheet of metal by cutting squares of equal size from the corners and folding up the sides. Determine the size of the squares to maximize the volume of the box.
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Let $x$ be the side length of the squares. We see that the length and width of the base will be $16-2x$ and $12-2x$, respectively. The height of the box will be $x$.
The volume is given by $V=x(16-2x)(12-2x)=4x^3-56x^2+192x$.
We will maximize this function over the interval $(0,6)$, as $x$ must be positive and $16-2x$ and $12-2x$ must also be positive (otherwise, there would be no base to the box).
To find the maximum, we see that the critical points occur when $V$ is undefined (when $x=0$) or when the derivative is 0.
Taking the derivative of $V$, we get $V'(x)=12x^2-112x+192$. Setting this equal to 0 yields $(3x-16)(4x-12)=0$. We find that $x=\frac{16}{3}, 3$.
Checking the values at these critical points, we find that $V\left(\frac{16}{3}\right)=\frac{2048}{9}$ and $V(3)=864$, so the maximum volume is $\boxed{\frac{2048}{9}}$.
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