Minimum value of $\sqrt{x^2+25}+\sqrt{y^2+16}$ if $x+y=12$

algebra-precalculus

$\sqrt{x^2+25}$ is the distance between $(x,0)$ and $(0,5)$, while $\sqrt{(x-12)^2+16}$ is the distance between $(x,0)$ and $(12,-4)$. Hence we need to find $x$ such that the sum of these two distances reaches the minimum. That is exactly the point that the line through $(12,-4)$ and $(0,5)$ meets the $x$-axis. Therefore we get $x = \frac{20}{3}$ and so $y = 12-\frac{20}{3} = \frac{16}{3}$.


We are asked for the minimum value of $\sqrt{x^2+25}+\sqrt{y^2+16}$ if $x+y=12;$

that is, the minimum value of $\sqrt{x^2+25}+\sqrt{(12-x)^2+16}.$

Let $\overrightarrow a=(x,5)$ and $\overrightarrow b=(12-x,4)$ in $\mathbb R^2,$ so $\overrightarrow a + \overrightarrow b=(12,9).$

By the triangle inequality, $|\overrightarrow a+\overrightarrow b|\le|\overrightarrow a|+|\overrightarrow b|.$

Therefore, $\mathbf{15}=\sqrt{12^2+9^2}\le\sqrt{x^2+25}+\sqrt{(12-x)^2+16}.$

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