10. It is required to fence a rectangular area adjacent to a wall. The fence should have a length of \(60 \mathrm{~m}\). What should be the length and width of this area so that its area is the largest?
10. Solution. First method. Let the width of the site be \(x \mu\), then its length will be ( \(60-2 x\) ) \(m\), and the area will be \(y=x(60-2 x)\) square meters. Considering the obtained equation as a quadratic equation in terms of \(x\), we get: \(2 x^{2}-60 x+y=0\). The discriminant of this equation is non-negative, therefore, \(900-2 y \geqslant 0, y \leqslant 450\). The maximum area is 450 square meters; the width of the site is 15 m, and the length is \(30 m\).

Second method. We can use the method of completing the square:

\[
\begin{aligned}
y=2\left(30 x-x^{2}\right)= & 2\left[225-\left(225-30 x+x^{2}\right)\right]=2[225- \\
& \left.-(15-x)^{2}\right]
\end{aligned}
\]

When \(x=15\), we get the maximum value of \(y\), which is 450.

Other methods of solution are possible. In particular, the problem can be solved based on geometric considerations.