15. Let $x, y$ be real numbers, then $\max _{5 x^{2}+4 y^{2}=10 x}\left(x^{2}+y^{2}\right)=$
4 Solution: $5 x^{2}+4 y^{2}=10 x \Rightarrow 4 y^{2}=10 x-5 x^{2} \geq 0 \Rightarrow 0 \leq x \leq 2$ $4\left(x^{2}+y^{2}\right)=10 x-x^{2}=25-(5-x)^{2} \leq 25-3^{2} \Rightarrow x^{2}+y^{2} \leq 4$