Task B-2.3. If $3 x+2 y=1$, determine the minimum value of the expression $x^{2}+y^{2}$.
## Solution.

From $3 x+2 y=1$ it follows that

$$
y=\frac{1-3 x}{2}
$$

Then,

$$
x^{2}+y^{2}=x^{2}+\left(\frac{1-3 x}{2}\right)^{2}=\frac{13}{4} x^{2}-\frac{3}{2} x+\frac{1}{4}
$$

is a quadratic function in $x$ which has a minimum

$$
y_{T}=\frac{4 \cdot \frac{13}{4} \cdot \frac{1}{4}-\left(-\frac{3}{2}\right)^{2}}{4 \cdot \frac{13}{4}}=\frac{1}{13}
$$

Therefore, $x^{2}+y^{2} \geq \frac{1}{13}$ (the smallest value of the expression is $\frac{1}{13}$ ).