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

If $$x,y\in\mathbb R^+$$, $$x+y=12$$, what is the minimum value of $$\sqrt{x^2+25}+\sqrt{y^2+16}$$?

I got the question from a mathematical olympiad competition (of China) for secondary 2 student, so I don't expect an "analysis" answer.

The answer should be $$15$$, when $$x=\frac{20}{3}$$ and $$y=\frac{16}{3}$$ (just answer, no solution :< ).

I try to use AM-GM inequality, but I couldn't manage to get the answer.

$$\sqrt{x^2+25}+\sqrt{y^2+16}=\sqrt{x^2+25}+\sqrt{(x-12)^2+16}=\sqrt{x^2+25}+\sqrt{x^2-24x+160}$$ Can this help?

I also tried to plot the graph, see here.

Any help would be appreciated. Thx!

• Note that $\left(\dfrac {20}3\right)^2=\left(\dfrac{20}3-12\right)^2+16$ – J. W. Tanner May 2 at 13:59
• $\sqrt{x^2+a^2}+\sqrt{(c-x)^2+b^2}\ge\sqrt{c^2+(a+b)^2}$; here $a=5, b=4,$ and $c=12$ – J. W. Tanner May 2 at 22:34

$$\sqrt{x^2+25}$$ 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 $$y = 12-\frac{20}{3} = \frac{16}{3}$$.

• Why $(0,5)$ and $(12,-4)$? It could be $(0, -5)$ or $(12,4)$ – Vasya May 2 at 14:06
• The same. Just find $(x,0)$ such that the sum of two distance reaches the minimum. We will get the same answer if we use these points. – Hongyi Huang May 2 at 14:08
• No, it is different if we use $(0, -5)$ and $(12,4)$ because we will never intersect axis $x$. – Vasya May 2 at 14:30
• Exactly the same, by symmetry.@Vasya – Hongyi Huang May 2 at 14:34
• The question is to find $x$ such that the sum reaches min. So we choose $(0,5)$ and $(12,-4)$ instead of $(0,5)$ and $(12,4)$ because we need the segment to intersect axis $x$, and thus the minimum is got. – Hongyi Huang May 2 at 14:37

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

• Equality holds when $\overrightarrow a \parallel \overrightarrow a + \overrightarrow b,$ i.e., $x/12=5/9$, i.e., $x=60/9=20/3$ – J. W. Tanner May 3 at 0:32
• Best solution on this page :-) – Andreas May 3 at 13:26

Here's an algebraic solution.

Let $$x = z + 20/3$$ then $$\sqrt{x^2+25}+\sqrt{y^2+16}=\sqrt{x^2+25}+\sqrt{(x-12)^2+16}\\ =\sqrt{(z+20/3)^2+25}+\sqrt{(z-16/3)^2+16}\\ = \sqrt{z^2 + 40 z/3 + 625/9}+\sqrt{z^2 - 32 z/3 + 400/9} = f(z)$$ Then $$f(z)^2= 1025/9 + 2 z^2 + 8z/3 + 2 \sqrt{z^2 + 40 z/3 + 625/9}\sqrt{z^2 - 32 z/3 + 400/9} \\ = 1025/9 + 2 z^2 + 8z/3 + 2 \sqrt{z^4 + (8 z^3)/3 - (85 z^2)/3 - (4000 z)/27 + 250000/81}$$ Since $$f(z)$$ is always positive, we can claim that $$f(z)^2 \ge 15^2 = 225$$ so we need to show that $$1025/9 + 2 z^2 + 8z/3 + 2 \sqrt{z^2 + 40 z/3 + 625/9}\sqrt{z^2 - 32 z/3 + 400/9} \ge 2025/9\\ \leftrightarrow \sqrt{z^4 + (8 z^3)/3 - (85 z^2)/3 - (4000 z)/27 + 250000/81} \ge 500/9 - z^2 - 4z/3 \\ \leftrightarrow z^4 + (8 z^3)/3 - (85 z^2)/3 - (4000 z)/27 + 250000/81 \ge (500/9 - z^2 - 4z/3)^2 = \\z^4 + (8 z^3)/3 - (328 z^2)/3 - (4000 z)/27 + 250000/81 \\ \leftrightarrow - (85 z^2)/3 \ge - (328 z^2)/3$$ and this is obviously true, with equality for $$z=0$$.

Let

$$f = \sqrt{x^2 + 25} + \sqrt{y^2 + 16}$$

$$g = x+y -12 =0$$

By Lagrange's Undetermined multipliers method,

$$F = f + g\lambda = \sqrt{x^2 + 25} + \sqrt{y^2 + 16} + \lambda(x+y -12)$$

Differentiating partially w.r.t x and y and using $$F_x = 0$$ and $$F_y = 0$$ at extremum,

$$F_x = \frac{x}{\sqrt{x^2 + 25}} +\lambda = 0$$

$$F_y = \frac{y}{\sqrt{y^2 + 16}} +\lambda = 0$$

So, $$\frac{x}{\sqrt{x^2 + 25}} = \frac{y}{\sqrt{y^2 + 16}}$$

$$\frac{x^2}{x^2 + 25} = \frac{y^2}{y^2 + 16}$$

$$\frac{x^2 + 25}{x^2} = \frac{y^2 + 16}{y^2}$$

$$1 + \frac{25}{x^2} = 1 + \frac{16}{y^2}$$

On solving,

$$5y = \pm 4x$$

Using this in g,

$$x \pm 4x/5 = 12$$

$$x = 20/3, 60$$

Using these values,

$$y = 12 - x$$

$$y = 12 -20/3 = 16/3$$ and $$y = 12 - 60 = -48$$

Using these values of x and y in f we find its value is minimum at $$(\frac{20}{3},\frac{16}{3})$$.

• I think OP wanted an answer without differentiation – J. W. Tanner May 2 at 14:18
• ... and the expression for $F_y$ is missing a $\lambda$. – NickD May 2 at 19:23
• Thank you Nick. – Ak19 May 3 at 0:44