Chapter 12 - Section 12.2 - Applications of Maxima and Minima - Exercises - Page 888: 41

$59.25 \ in^3$

We have: $V(x)=(16-2x)(6-2x)x=4x^3-44x^2+96x$ For maximum volume, we must have $\dfrac{dV}{dx}=0$ or, $12x^2-88x+96=0$ or, $3x^2-22x+24=0$ ,or, $x=\dfrac{22 \pm \sqrt{484-288}}{6}=\dfrac{4}{3} \ in.$ Thus, the maximum volume is: $V(x)=4(\dfrac{4}{3})^3-44(\dfrac{4}{3})^2+96\times \dfrac{4}{3}=59.25 \ in^3$