jmc

The graph of $y=4(x+7)(2-x)$ is a parabola. Since $y=0$ when $x=-7$ and $x=2$, the $x$-intercepts of the parabola are $(-7,0)$ and $(2,0)$. If the vertex of the parabola is $(h,k)$, then the $x$-intercepts $(-7,0)$ and $(2,0)$ are symmetric around the line $x=h$, so $h=(-7+2)/2=-5/2$.

Hence, the maximum value of $y=4(x+7)(2-x)$ occurs at $x=-5/2$, in which case $$y=4\left(-\frac{5}{2}+7\right)\left(2+\frac{5}{2}\right)=4\cdot \frac{9}{2}\cdot \frac{9}{2}=\boxed{81}.$$ (Note that this is a maximum value, and not a minimum value, because the coefficient of $x^2$ in $y=4(x+7)(2-x)=-4x^2-20x+56$ is negative.)