Optimisation Problem on Cone

The problem I’ve got here is to prove that semi vertical angle of a cone with maximum volume with total surface area constant is equal to $arcsin(\frac{1}{3})$
I am trying to do that by making some relation in terms of trigonometric equations and then differentiating it.
Supposing slant height is $l$ and semi vertical angle is $\theta$
then radius becomes $r=l\sin(\theta)$ and height becomes $h=l\cos(\theta)$
Given
$$S=\pi r^2+\pi r l=constant$$
$$V=\frac{1}{3}\pi r^2h$$
but somehow i am not able to get the exact relation between
$$V=\frac{1}{3}l^2\cos^2(\theta).l\sin(\theta)$$
then I am taking Derivative of Function $V(\theta)$ and equating to zero to get maxima , but it’s not working out , help ?

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$$\pi r(r+l) = k$$
$$l^{2}(1+sin\theta)sin\theta = k$$

$V=\frac{1}{3}l^{2}sin^{2}\theta lcos\theta$ This is where you made a mistake.

Substitute the value of $l^2$ and $l$ in terms of $\theta$ in V and use the identity
$$cos\theta = \sqrt{(1-sin\theta)(1+sin\theta)}$$

$$ = k\dfrac {\sqrt{sin\theta(1-sin\theta)}}{(1+sin\theta)}$$

Put $u = sin\theta$

$$V = K\dfrac{\sqrt{u(1-u)}}{1+u}$$

Set $$\frac{dV}{du} = 0$$

And using quotient rule, you will get $$3u-1=0$$

$$ u = \frac{1}{3}$$

$$sin\theta = \frac{1}{3}$$

$$ \theta = arcsin(\frac{1}{3})$$