Differentiating equations

Question

I'm an A-level student having trouble with approaching this question, looking at the mark scheme also gives no real tips on how to approach it, please help if possible.

Edit:

So when starting a) I looked at the equations and saw $h = \sqrt{l^2 -r^2}$, and $l=4/r$. I subbed l into the equation for height and got $h = \sqrt{(16/r^2) - r^2}$

...WAIT I think I got it! I just sub $h$ into the volume equation!

yeah if i do that i get $V = 1/3 \pi r^2 \sqrt{(16/r^2) - r^2}$, which simplifies to

$\pi/3 \sqrt{(16/r^2)(r^4) - r^2(r^4)}$

which simplifies to

$\pi/3 \sqrt{16r^2 - r^6}$

WHICH IS THE ANSWER!!

now for b):

since it's asking for a maximum and is asking for a relation between V and r is integration/differentiation but I don't know which one, nvm I should differentiate, when I do that I get:

i can't im getting stuck i feel dumb, im at $1/3 \pi (16r^2 -r^6)^{1/2}$

i forgot how to differentiate that, do i do every term e.g. $1/6 \pi (32r^1 -6r^5)^{-1/2}$

or something else?

edit:

Okay after doing a confusing differentiation i got to

$(\pi(32-6r^5))/(6\sqrt{16r^2 -r^6}) = 0$ for the maximum that simplifies to $32-6r^5 = 0$ right?

then $6r^5 = 32$ $r = \sqrt[5]{32/6} = 1.40$ to $3$S.F but then the mark scheme says $1.52...$

EDIT: AGAIN i made a nooby mistake

I'm at the answer of r = $1.52$

because of me not square rooting it by $4$ instead of $5.$

now I sub r into the $\pi/3 \sqrt{16r^2 - r^6}$

and got $V_{max} = 5.197 == 5.2$ WHICH IS RIGHT!!!

FINALLY :D

Answer 1

There is a relationship between $r,h,$ and $l$! Use that, and use the other formulae.

That should solve part (a).

To solve part (b), differentiate $V=V(r)$ and find where it is zero $\cdots.$

Answer 2

As it is a symmetric cone , it follows that $h^2=l^2-r^2$, where $l$ is slant height , $h$ is height and $r$ is radius of base . Also $rl=4.$ Therefor $l=4/r$. Now part a) can be easily solved .

For part b) , when $V$ is maximum , $V^2$ is also maximum , therefore $d(V^2)/dr = 0$. Therefore $d(16r²-r^6)/dr = 0.$ Solve and you will get the value of $r$ at which $V$ is maximum very easily .