Local Extremas

Let x represent the side length of the square. The volume will be $(16-2x)(13-2x)(x)$ Since each time you increase the side length by 1, the height of the box increases 1, the length and width decreases 2. Expanding this equation gets you a polynomial function: $f(x)=4x^{3}-58x^{2}+208x$ You want to find the local maxima of this function. Since when you differentiate a polynomial function, its local extremas became zeroes, you can differentiate $f(x)$ using the power rule. $\frac{d}{dx}f(x)$ $=\frac{d}{dx}4x^{3}-\frac{d}{dx}58x^{2}+\frac{d}{dx}208x$ $=3 \times 4x^{(3-1)}-2 \times 58x^{(2-1)}+1 \times 208x^{(1-1)}$ $f'(x)=12x^{2}-116x+208$ Applying the quadratic formula on this function gives you the 2 extremas of the original function. $x=\frac{-b+\sqrt{b^{2}-4ac}}{2a}$ $x=\frac{-b-\sqrt{b^{2}-4ac}}{2a}$ which means $x=2.3782, x=7.2885$ Now, check if those 2 answers are valid to the original question. Since a box can not have negative width, 7.2885 is not possible ( $13-2 \times 7.2885<0$ ). Substituting 2.3782 into $f(x)$ gets you the maxima 220.4299. That means when the side length of the squares cutted out is 2.3782, the volume is the biggest, which is 220.4299. Adding 2.3782 and 220.4299 gives you the correct answer: 222.81.