Minimum Distance Problem 2
For $ 0 \le x \le 40 $ and $ 0 \le y \le 50, $ find the minimum value of
\[\sqrt{x^2 + 400} + \sqrt{y^2 + 900} + \sqrt{x^2 + y^2 - 80x - 100y + 4100}.\]
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Solution
Completing the square in $ x $ and $ y, $ the expression becomes
\[\sqrt{x^2 + 400} + \sqrt{y^2 + 900} + \sqrt{(x - 40)^2 + (y - 50)^2} = \sqrt{x^2 + 400} + \sqrt{y^2 + 900} + \sqrt{(40 - x)^2 + (50 - y)^2}.\]By QM-AM,
\begin{align*}
\sqrt{\frac{x^2 + 400}{2}} &\ge \frac{x + 20}{2}, \\
\sqrt{\frac{y^2 + 900}{2}} &\ge \frac{y + 30}{2}, \\
\sqrt{\frac{(40 - x)^2 + (50 - y)^2}{2}} &\ge \frac{(40 - x) + (50 - y)}{2},
\end{align*}so
\begin{align*}
&\sqrt{x^2 + 400} + \sqrt{y^2 + 900} + \sqrt{(40 - x)^2 + (50 - y)^2} \\
&\ge \sqrt{2} \cdot \frac{x + 20}{2} + \sqrt{2} \cdot \frac{y + 30}{2} + \sqrt{2} \cdot \frac{(40 - x) + (50 - y)}{2} \\
&= 70 \sqrt{2}.\end{align*}Equality occurs when $ x = 20 $ and $ y = 30, $ so the minimum value is $ \boxed{70 \sqrt{2}} $.
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