# A huge conical tank to be made from a circular piece of sheet metal of radius 10m by cutting out a s

A huge conical tank to be made from a circular piece of sheet metal of radius 10m by cutting out a sector with vertex angle theta and welding the straight edges of the straight edges of the remaining piece together. Find theta so that the resulting cone has the largest possible volume.

Specifically, the question is asked in the context of wanting derivatives, multiple max/min equations, and hopefully more calc rather than trig or geo.

I have gotten as far as using 10m as the hypotenuse for a triangle formed by the height of the cone, radius of the base of the cone, and slant. I'm not sure where to go from there, because I can't determine how to find height and/or radius, without which I'm not sure I can continue.
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Haylie Cherry
The requested volume will be (why?):
$V\left(\theta \right)=\pi {R}^{2}\frac{h}{3}=\pi {\left(\frac{2\pi -\theta }{2\pi }r\right)}^{2}\sqrt{{r}^{2}-{\left(\frac{2\pi -\theta }{2\pi }r\right)}^{2}}$
In here, the given variable is $r=10$, and the unknown variable $\theta$ must be found, by taking the derivative $\frac{dV\left(\theta \right)}{d\theta }$.

You can change the variable and analyze better for the variable $R=\frac{2\pi -\theta }{2\pi }r$, which will be much easier:
$V\left(R\right)=\pi {R}^{2}\sqrt{{r}^{2}-{R}^{2}}$
You will find the optimal is given for $R=\sqrt{\frac{2}{3}}r$, or $\theta =2\pi \left(1-\sqrt{\frac{2}{3}}\right)={66.1}^{o}$