# Find the minimum value of $\sqrt{x^2+y^2-4x+2y+5}+\sqrt{x^2+y^2+6x-4y+13}$ without calculus.

by Rogério   Last Updated September 23, 2018 00:20 AM

Is possible to find the minimum value of

$$f(x,y)=\sqrt{x^2+y^2-4x+2y+5}+\sqrt{x^2+y^2+6x-4y+13}$$

without calculus?

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Of course.

$$f(x,y)=\sqrt{(x-2)^2+(y+1)^2}+\sqrt{(x+3)^2+(y-2)^2}.$$

Now define the vectors

$$\vec{a}=(x-2,y+1) \to |\vec{a}|=\sqrt{(x-2)^2+(y+1)^2},$$ $$\vec{b}=(x+3,y-2) \to |\vec{b}|=\sqrt{(x+3)^2+(y-2)^2},$$

such that,

$$f(x,y)=|\vec{a}|+|\vec{b}|.$$

Looking at the following well-known inequality

$$|\vec{a}|+|\vec{b}|\geq|\vec{a}-\vec{b}|,$$

implies that

$$f(x,y)\geq\sqrt{34}.$$

We must now prove that there are $$x_1$$ and $$y_1$$ such that $$f(x,y)=\sqrt{34}.$$

Note that if

$$f(x,y)=\sqrt{34}\to|\vec{a}|+|\vec{b}|=|\vec{a}-\vec{b}|.$$

Then the vectors $$\vec{a}$$ and $$\vec{b}$$ are collinear. Thereby,

$$\frac{x-2}{x-3}=\frac{y+1}{y-2} \to x=\frac{1-5y}{3}.$$

If $$y_1=-1$$ we get $$x_1=2$$, such that $$f(2,-1)=\sqrt{34}$$.

Therefore the minimum value of $$f(x,y)$$ is

$$f_{min}(x,y)=\sqrt{34}.$$

Dinesh Shankar
September 22, 2018 23:39 PM