Question

A conical tent of given capacity has to be constructed. The ratio of the height to the radius of the base for the minimum amount of canvas required for the tent is:

• A. 1 : 2
• B. 2 : 1
• C. \(1 : \sqrt2\)
• D. \(\sqrt2 : 1\)

Hint:

Fix the volume, write the curved surface area as a function of one variable, either r or the ratio h/r, and minimise it using calculus.

Answer 1

The Correct Option is D

Step 1: Set up the known quantities.
Let the cone have base radius \(r\), height \(h\), and slant height \(l = \sqrt{r^2+h^2}\). The volume, or capacity, of the tent is fixed, say \(V = \frac{1}{3}\pi r^2 h\). Since a tent has no floor or top, the canvas needed is just the curved surface area \(S = \pi r l = \pi r\sqrt{r^2+h^2}\). We want the ratio \(h:r\) that minimises \(S\) while \(V\) stays fixed.

Step 2: Express h in terms of r using the fixed volume.
From \(V = \frac{1}{3}\pi r^2h\), we get \(h = \frac{3V}{\pi r^2}\).

Step 3: Write S purely in terms of r, and work with S squared to avoid the square root.
\[ S^2 = \pi^2 r^2 (r^2+h^2) = \pi^2 r^4 + \pi^2 r^2 h^2 \] Since \(\pi r^2 h = 3V\), squaring gives \(\pi^2 r^4 h^2 = 9V^2\), so \(\pi^2 r^2 h^2 = \frac{9V^2}{r^2}\). Substituting:
\[ S^2 = \pi^2 r^4 + \frac{9V^2}{r^2} \]
Step 4: Minimise S squared with respect to r.
Differentiate with respect to \(r\) and set the derivative to zero:
\[ \frac{d(S^2)}{dr} = 4\pi^2 r^3 - \frac{18V^2}{r^3} = 0 \implies 4\pi^2 r^6 = 18V^2 \implies r^6 = \frac{9V^2}{2\pi^2} \]
Step 5: Bring back h to find the ratio.
Recall \(V = \frac{1}{3}\pi r^2 h\), so \(V^2 = \frac{1}{9}\pi^2 r^4 h^2\). Substitute into the relation above:
\[ r^6 = \frac{9}{2\pi^2}\left(\frac{1}{9}\pi^2 r^4h^2\right) = \frac{r^4h^2}{2} \] Dividing both sides by \(r^4\):
\[ r^2 = \frac{h^2}{2} \implies h^2 = 2r^2 \implies \frac{h}{r} = \sqrt2 \]
Step 6: Confirm this is a minimum, and check the other options.
As \(r \to 0\) or \(r \to \infty\), \(S \to \infty\), since the tent becomes either a very tall thin spike or a very flat wide cone, both needing a huge amount of canvas, so the single critical point found is a minimum, not a maximum. Options (A) 1:2, (B) 2:1, and (C) \(1:\sqrt2\) do not satisfy \(h^2=2r^2\), so they do not correspond to the actual minimum.

Final Answer:
The ratio of height to radius for minimum canvas is \(\sqrt2:1\). \[ \boxed{h:r = \sqrt2 : 1} \]