The maximum volume of the box is required.
The height of the box is 1.81 ft
The maximum volume of the box is \(24.19\ \text{ft}^3\).
The given function is
\(V(h)=h(h-5)(h-6)\\\Rightarrow V(h)=h^3-11h^2+30h\)
Differentiating with respect to \(h\)
\(V'(h)=3h^2-22h+30\)
Equating to zero
\(0=3h^2-22h+30\\\Rightarrow h=\frac{-\left(-22\right)\pm \sqrt{\left(-22\right)^2-4\times3\times30}}{2\times3}\\\Rightarrow h=5.52,1.81\)
Double derivative of the function
\(V''(h)=6h-22\)
Substituting \(h\) values
\(V''(5.52)=6\times 5.52-22=11.22\)
\(V''(1.81)=6\times 1.81-22=-11.14\)
Since, \(V''(1.81)\) is negative, the function will be maximum at \(h=1.81\).
Checking domain \(1.81\in (0,6)\)
The maximum volume is
\(V(1.81)=1.81^3-11\times 1.81^2+30\times 1.81=24.19\ \text{ft}^3\)
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