Answer:
[tex]{ \rm{V(x) = 4x {(10 - x)}^{2} }} \\ \\ { \rm{ \frac{dv}{dx} = 4 {(10 - x)}^{2} + 4x( - 2(10 - x))}} \\ \\ { \rm{ \frac{dv}{dx} = 4 {(10 - x)}^{2} + 4x( - 20 - 2x) }} \\ \\ { \rm{ \frac{dv}{dx} = 4(100 - 20x + {x}^{2} ) + ( - 80x - 8 {x}^{2} ) }} \\ \\ { \rm{ \frac{dv}{dx} = - 4 {x}^{2} - 160x + 400}}[/tex]
• But at maximum volume, dv/dx is 20 square inches.
[tex]{ \rm{ - 4 {x}^{2} - 160x + 400 =2 0}} \\ \\ { \rm{ {x}^{2} + 40x - 95 = 0}}[/tex]
• Using the quadratic formula:
[tex]{ \underline{ \rm{ \: \: x \approx3\: \: squares \: \: }}}[/tex]
• Therefore,
For maximum volume,
[tex]{ \rm{V(x) = 4x(10 - x) {}^{2} }} \\ \\ { \rm{V(3) = (4 \times 3)(10 - 3) {}^{2} }} \\ \\ { \rm{V(3) = 12 \times 49}} \\ \\ { \boxed{ \rm{V = 588 \: {in}^{3} }}}[/tex]
