What is the maximum value of $H(r)=\frac{1}{3} \pi r^2\sqrt{l-r^2} $ when $l=7$?

Question

What is the maximum value of $H(r)=\frac{1}{3} \pi r^2\sqrt{l-r^2} $ when $l=7$?

I've computed the minima using derivative method,but I'm not able to calculate the maxima under the given condition...

Please suggest me how can i Calculate the maxima...

Answer 1

As Lord Shark the Unknown has mentioned in the comment, let $G(r) = r^4(7-r^2)$

So, $G'(r) = 28r^3-6r^5 , G''(r) = 84r^2-30r^4$

At maxima, $G'(r) = 0 \text{ and } G''(r) <0$

$G'(r) = 0 \implies r^2 = \frac{28}{6} \implies r = \pm \sqrt{\frac{14}{3}} $

$G''(r)= 84\times \frac{14}{3} -30\times \frac{196}{9} < 0$

So

$\color{darkgreen}{r=\pm \sqrt{\frac{14}{3}}}$