Proof: minimum volume of a notch cut at equal angle of cutting surfaces with horizontal plane > A notch is cut in a cylindrical vertical tree trunk. The edge of the cut reaches the axis of the cylinder and the cut is between two half-circle planes. Each half-circle is bounded by a horizontal line passing through the axis of the cylinder. The angle between the two half-circle planes is θ. Prove that the volume of the notch is minimized (for given tree and θ) by taking the half-circle planes at equal angles with the horizontal plane. I'm having trouble even visualizing the problem. I think it would be useful to find the total volume of the notch in terms of the angles that the bounding planes form with the horizontal plane and then differentiate to find when the minimum is achieved. Can someone provide a solution?

Let $r$ be the trunk radius, and $\alpha$ and $\beta$ the angles formed by the two cut-planes with the horizontal plane, thus $\alpha+\beta=\theta$.

The equation for the trunk cylinder can be written as $x^2+y^2=r^2$, and the two cut-planes as $z_1=y\tan\alpha$ and $z_2=-y\tan\beta$. Then, integrate $z_1-z_2$ over a half-circle in the $xy$-plane to obtain its volume,

$$V = \int_{-r}^{r}\int_0^{\sqrt{r^2-x^2}}[y\tan\alpha -(-y\tan\beta)] \>dydx=\frac23 r^3(\tan\alpha+\tan\beta)$$

Next, with $\beta=\theta-\alpha$, take the derivative with respect to $\alpha$ and set it to zero to minimize $V$,

$$\sec^2\alpha - \sec^2\beta=0$$

which yields $\alpha = \beta$. Hence, the notch volume is minimal at equal angles of the cutting surfaces with the horizontal plane.