Solve for m_2
m_{2}=\sqrt{800-5y}+\sqrt{5y}
y\geq 0\text{ and }y\leq 160
Solve for y
y=\frac{m_{2}\sqrt{1600-m_{2}^{2}}}{10}+80
y=-\frac{m_{2}\sqrt{1600-m_{2}^{2}}}{10}+80\text{, }m_{2}=20\sqrt{2}\text{ or }\left(m_{2}\leq 40\text{ and }m_{2}\geq 20\sqrt{2}\text{ and }m_{2}\sqrt{1600-m_{2}^{2}}\leq 800\right)
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