What Are the Dimensions of a Canvas Tent for Maximum Volume?

[roam]

Hello!

Here's a question that I couldn't understand;

A canvas tent is to be constructed in the shape of a right-circular cone with the ground as base;

Using the volume V and curved surface area S of the cone,
$$V = \frac{1}{3}\pi r^2 h$$, $$S = \pi rl$$,

Find the dimensions of the cone that maximises the volume for the $$4 \sqrt{3 \pi}$$ m² of canvas material, and find this maximum volume.


I'd appreciate it if you could give me some hints and guidance so I can get started on this question. I really don't understand what this question means, and there are no measures of height, base, redius etc.. are given...


Thank you.

 

[HallsofIvy]

Hello!

Here's a question that I couldn't understand;

A canvas tent is to be constructed in the shape of a right-circular cone with the ground as base;

Using the volume V and curved surface area S of the cone,
$$V = \frac{1}{3}\pi r^2 h$$, $$S = \pi rl$$,

Find the dimensions of the cone that maximises the volume for the $$4 \sqrt{3 \pi}$$ m² of canvas material, and find this maximum volume.


I'd appreciate it if you could give me some hints and guidance so I can get started on this question. I really don't understand what this question means, and there are no measures of height, base, redius etc.. are given...


Thank you.

Yes, those are what you are asked to find! It does say "find the dimensions". The problem is asking: out of all cones with surface area $4\sqrt{3 \pi}$ m^[2 of material, which r and h (radius and height) will give the largest volume? You know the two formulas: $V= \pi r^2h/3$, $S= \pi rl$. l, of course, is the "slant height". Using the Pythagorean theorem, l²= r²+ h².

As to how to solve the problem, the best method depends upon your level of study. Since you don't seem to have seen problems like this you are probably in basic calculus. Since you must have $S= \pi rl= \pi r\sqrt{r^2+ h^2}= 4\sqrt{3\pi}$. Squaring both sides of that $\pi^2r^2(r^2+ h^2)= 48\pi$. You can solve that for h and replace h in V= \pi r^2h/3 by that to get a problem in just the one variable r. Use whatever methods you know to find the largest possible value of that function.