## Task 30/80

A pentagon consists of a rectangle with sides $a$ and $b$ and an equilateral triangle with side $a$.

Without using differential calculus, determine the maximum area $A$ for a given perimeter $U$ and the minimum perimeter $U$ for a given area $A$!
It is $U=3 a+3 b$ and $A=a b+0.25 a^{2} \sqrt{3}$. Eliminating the side $b$ from these equations, one obtains (after equivalent transformations) the quadratic equation in $a$

$$
a^{2}-\frac{2 U}{6-\sqrt{3}} a+\frac{4 A}{6-\sqrt{3}}=0
$$

with the solutions

$$
x_{1 ; 2}=\frac{1}{6-\sqrt{3}}\left[U \pm \sqrt{U^{2}-4 A(6-\sqrt{3}}\right]
$$

For $a$ to be real, it must be that

$$
U^{2}-4 A(6-\sqrt{3}) \geq 0 \rightarrow A \leq \frac{U^{2}}{4(6-\sqrt{3}} \rightarrow U \geq \sqrt{4 A(6-\sqrt{3})}
$$

Obviously, this implies

$$
A_{\max }=\frac{U^{2}}{4(6-\sqrt{3})} \approx 0.0586 U^{2} \quad ; \quad U_{\min }=\sqrt{4 A(6-\sqrt{3})} \approx 4.13 \sqrt{A}
$$