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:

Question

What is the value of $ \int_0^{\pi/2} \sin x \cos x \, dx$?

Answer 1

Here is another way to solve this, by working with the ratio $k = h/r$ directly instead of minimising in terms of $r$ first.

Let $k = \frac{h}{r}$, so $h = kr$. The volume is fixed: $V = \frac{1}{3}\pi r^2 h = \frac{1}{3}\pi r^3 k$, which gives $r^3 = \frac{3V}{\pi k}$.

The curved surface area is $S = \pi r l = \pi r \sqrt{r^2+h^2} = \pi r^2\sqrt{1+k^2}$, using $h=kr$.

Now express $r^2$ in terms of $k$: from $r^3 = \frac{3V}{\pi k}$, we get $r = \left(\frac{3V}{\pi k}\right)^{1/3}$, so $r^2 = \left(\frac{3V}{\pi k}\right)^{2/3}$.

Substituting into $S$:

$S = \pi \left(\frac{3V}{\pi k}\right)^{2/3}\sqrt{1+k^2}$.
Since $V$ is fixed, minimising $S$ over $k$ is the same as minimising $f(k) = k^{-2/3}\sqrt{1+k^2}$, because the other factors involving $V$ and $\pi$ do not depend on $k$.
Write $f(k) = k^{-2/3}(1+k^2)^{1/2}$ and take the logarithm to make differentiation easier: $\ln f(k) = -\frac{2}{3}\ln k + \frac{1}{2}\ln(1+k^2)$.

Differentiate both sides with respect to $k$, using logarithmic differentiation, $\frac{f'(k)}{f(k)} = \frac{d}{dk}[\ln f(k)]$:

\[ \frac{f'(k)}{f(k)} = -\frac{2}{3k} + \frac{k}{1+k^2} \]

Set this to zero, since $f(k) \neq 0$, the bracket must vanish:

\[ -\frac{2}{3k} + \frac{k}{1+k^2} = 0 \implies \frac{k}{1+k^2} = \frac{2}{3k} \implies 3k^2 = 2(1+k^2) \implies 3k^2 - 2k^2 = 2 \implies k^2 = 2 \]

So $k = \sqrt2$, taking the positive root since $k = h/r$ must be positive.

Let's summarize:

Working directly with the ratio $k = h/r$ turns this into a one-variable minimisation.
Logarithmic differentiation avoids messy product and chain rule steps.
Both this method and the direct r-based method agree: $k = h/r = \sqrt2$.

So the ratio of height to radius is $\sqrt2 : 1$, confirming option (D).

Answer 2