## Task 26/66

An area consisting of a rectangle with sides $a$ and $b$ and a semicircle with radius $r=\frac{b}{2}$ has an area $A=100 \mathrm{~cm}^{2}$.

Without using differential calculus, determine the sides $a$ and $b=2 r$ such that the perimeter of the area is minimized. What is $U_{\min }$?

(The semicircle borders the side $b$ of the rectangle with its straight boundary line.)
It is $A=a b+\frac{\pi}{8} b^{2}$ and $U=2 a+b+\frac{\pi}{2} b$. By eliminating the variable $a$ from these equations, one obtains the quadratic equation for the variable $b$

$$
b^{2}-\frac{4 U}{4+\pi} b+\frac{8 A}{4+\pi}=0
$$

The condition for the existence of a real solution is

$$
D=\frac{4 U^{2}-8 A(4+\pi)}{(4+\pi)^{2}} \geq 0
$$

From this follows $U^{2} \geq 2 A(4+\pi)$ and $U_{\min }=\sqrt{2 A(4+\pi)} \approx 37.8 \text{~cm}$ (slide rule accuracy). The quadratic equation for $b$ then has the double solution (correspondingly for $a$):

$$
b=\frac{2 U_{\min }}{4+\pi}=2 \sqrt{\frac{2 A}{4+\pi}} \approx 10.58 \text{~cm} ; \quad a=\frac{U_{\min }}{4+\pi}=\sqrt{\frac{2 A}{4+\pi}} \approx 5.29 \text{~cm}
$$

Note: One could have also eliminated $b$ initially and thus obtained a quadratic equation for $a$, which would have led to the same result.