## Task B-1.5.

Determine the minimum value of the expression $2 x^{2}+\frac{1}{2} y^{2}$ if $y+2 x=2$.
## Solution.

From the second equality, express $y=2-2 x$ and substitute into the first. Then we have

$$
\begin{aligned}
2 x^{2}+\frac{1}{2}(2-2 x)^{2} & =2 x^{2}+\frac{1}{2}\left(4-8 x+4 x^{2}\right) \\
& =4 x^{2}-4 x+2 \\
& =4 x^{2}-4 x+1+1 \\
& =(2 x-1)^{2}+1
\end{aligned}
$$

Since $(2 x-1)^{2} \geqslant 0$ for all $x \in \mathbb{R}$, the expression $(2 x-1)^{2}+1$ is greater than or equal to 1 for all real numbers $x$, so its minimum value is equal to 1.