## Task A-2.1.

Determine the radius of the base of a cone whose slant height is 1, so that the difference between the area of its lateral surface and the area of its base is maximized.
## Solution.

Let $r$ be the radius of the base of the cone, and let the lateral surface of the cone be a circular sector with radius 1 and angle $\alpha$.

Since the circumference of the base coincides with the circular arc over the angle $\alpha$, it follows that

$$
2 r \pi=\pi \frac{\alpha}{180^{\circ}}
$$

The area of the base is $B=r^{2} \pi$, and the area of the lateral surface is

$$
P=\pi \cdot \frac{\alpha}{360^{\circ}}=r \pi
$$

The desired difference in areas is $P-B=\pi\left(r-r^{2}\right)$, which is a quadratic function of $r$. It attains its maximum at the abscissa of the vertex, i.e., for $r=\frac{1}{2}$. In this case, the angle of the circular sector is $\alpha=180^{\circ}$. Therefore, the radius of the base for which the difference in the area of the lateral surface and the base is maximal is $\frac{1}{2}$.