## Task 30/62

To be investigated are frustums of cones with the same height and equal-area axial sections. How large must the radius of the top circle be so that the volume is maximized?

Solution a with the help of, b without the use of differential calculus.
For the volume $R$ of a frustum of a cone, the following applies:

$$
V=\frac{\pi h}{3}\left(R^{2}+R \cdot r+r^{2}\right)
$$

In the present case, the side conditions are: $h=$ constant (same heights), $\frac{R+r}{2} \cdot h=$ constant (equal cross-sectional areas). From the last equation, it follows that $R+r=constant=a$ or $R=a-r=$ $a-x$ (I) with $r=x$. Substituting (II) into (I) yields

$$
V=\frac{\pi}{3} h\left(a^{2}-2 a x+x^{2}+a x-x^{2}+x^{2}\right)=\frac{\pi}{3} h\left(x^{2}-a x+a^{2}\right)
$$

Solution a: The extreme values of $V$ are obtained by setting $V'=0$ and checking the value of $V''$:

$$
V'=\frac{\pi}{3} h(2 x-a) \quad V''=\frac{2 \pi}{3} h
$$

It is evident that $V''>0$ for any $x$, meaning the function $V$ has no relative maxima in the interior of the relevant interval. If maxima exist, they must lie at the interval boundaries.

From $R \geq 0$ and $r \geq 0$, it follows from (II) that $0 \leq x \leq a$ and $a \geq 0$. For $x_{1}=0$ and for $x_{3}=a$, we get $V=\frac{\pi}{3} h a^{2}$. For $0<\bar{x}<a$, $x^{2}<a x$, thus $x^{2}-a x<0$ and consequently $a^{2}-a x+a^{2}<a^{2}$, meaning that the maximum values indeed lie at the interval boundaries.

Solution b: From (III), it follows by simple rearrangement that

$$
V=\frac{\pi}{3} h x^{2}-\frac{\pi}{3} h a x+\frac{\pi}{3} h a^{2}
$$

It is evident that $V$ is a polynomial function of the second degree. Due to the positive coefficient of the quadratic term, the parabola representing the function opens upwards, meaning that maxima can only occur at the boundaries of the relevant interval.

From $R \geq 0$ and $r \geq 0$, it follows from (II) that $0 \leq x \leq a$, so that $x_{1}=0$ and $x_{2}=a$ yield a maximum volume: $V_{\max }=\frac{\pi}{3} h a^{2}$.

Conclusion 1: From (II), it follows that with $x_{1}=0, R=a$ and with $x_{2}=a, R=0$. This means that the volume is maximal when the frustum becomes a cone.

Conclusion 2: Not always does differential calculus offer a solution path in extremum value calculations.