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.)