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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.

jw35174
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  • Hint $\theta$ Is the ratio of circumference of the base to the slant height of the cone... Can you express the volume as a function of $\theta$? – Macavity Dec 05 '17 at 02:17

1 Answers1

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The requested volume will be (why?): $$ V(\theta)=\pi R^2 \frac h3 = \pi \left( \frac{2\pi-\theta}{2\pi}r \right)^2 \sqrt{r^2-\left(\frac{2\pi-\theta}{2\pi}r \right)^2} $$

In here, the given variable is $r=10$, and the unknown variable $\theta$ must be found, by taking the derivative $\frac{dV(\theta)}{d\theta}$.

You can change the variable and analyze better for the variable $R=\frac{2\pi-\theta}{2\pi}r$, which will be much easier: $$ V(R)= \pi R^2 \sqrt{r^2-R^2} $$

You will find the optimal is given for $R=\sqrt{\frac23}r$, or $\theta=2\pi(1-\sqrt{\frac23})=66.1^o$

Brethlosze
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  • My question now is how to derive the first equation(s) with the given information. I feel like I would understand the information better, but I can't jump the steps to how you got there. Could you continue the process? – jw35174 Dec 05 '17 at 02:52
  • I think that part is your homework.... dont you think? It is just geometry. You have the original circle of radius $r$ and perimeter $2\pi r$, which you take a sector away, keeping (of course) the radius $r$ but reducing the perimeter to $(2\pi-\theta)r$. Clear? – Brethlosze Dec 05 '17 at 02:56
  • Hence, when you curve the cone, you form a new circle of radius $R$ smaller, and perimeter $(2\pi-\theta)r=2\pi R$. Hence you have the equation of $R$ for your new circle. The height of the cone is by Pithagoras $\sqrt{r^2-R^2}$. Right? – Brethlosze Dec 05 '17 at 02:58