Question

How to solve V(h)=h(h-10)(h-8) find maximum for the domain 0 13 years ago

Answer 1

Take the derivative with respect to h, first, realize that for the product rule of 3 functions we have:
\[(fgh)'=f'gh+fg'h+fgh'\]
An extension of the 2 factor one.
So we have:
\[\frac{d V}{dh}=(1)(h-10)(h-8)+h(1)(h-8)+h(h-10)(1)=0\]
\[\implies (h-10)((h-8)+h)+(h-8)h=0 \implies h^2+80-18h+h^2-10h+h^2-8h=0\]
\[\implies h^2+80-18h+h^2-10h+h^2-8h=0 \]
\[\implies 3h^2-36h+80=0 \implies h=\frac{36 \pm \sqrt{36^2-4*3*80}}{6} \implies h=\frac{2}{3}(9 \pm \sqrt{21})\]
Both are within 0<h<10

Answer 2

So then take the second derivative and test the concavity at these points to see if its a max or min