What's the maximum volume for this cone?

Question

What's the maximum volume a right circular cone with a slant height of 45 length units possible can have?

The volume $V$ for a right circular cone is given by $V = \frac{\pi r^2 h}{3}$.

Answer 1

Hint

Starting from lab bhattacharjee's answer, $$V(h)=\dfrac{\pi}3h(45^2-h^2)$$ and you want to maximize $V$.

When is reached an extremum of a function ? So, since you know it, write the corresponding condition and solve for $h$. Plug this value in $V(h)$ and use the second derivative test to confirm that the result corresponds to a maximum.

I am sure that you can take from here.

Answer 2

We have $r^2+h^2=45^2\iff r^2=\cdots$

$$V=\dfrac{\pi r^2h}3=\dfrac{\pi}3h(45^2-h^2)=f(h)$$

Apply Second Derivative Test on $f(h)$

Or

$$\dfrac{ah+b(45-h)+c(45+h)}{1+1+1}\ge\sqrt[1+1+1]{ah\cdot b(45-h)\cdot c(45+h)}$$

WLOG set $ a-b+c=0\iff b=c+a$ so that the left hand side becomes a constant

We need $ah=c(45+h)=(c+a)(45-h)$ for the equality

$$\implies\dfrac ca=\dfrac h{45+h}$$ and $$\dfrac h{45-h}=\dfrac{c+a}a=1+\dfrac ca$$

Comparing the two values of $\dfrac ca, 3h^2=45^2$