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QUESTION: Let $ƒ(x,y)=\sqrt{x^2+y^2}+\sqrt{x^2+y^2-2x+1}+\sqrt{x^2+y^2-2y+1}+\sqrt{x^2+y^2-6x-8y+25}$

(A) Minimum value of $ƒ(x,y)= 5+\sqrt2$

(B) Minimum value of $ƒ(x,y)= 5-\sqrt2$

(C) Minimum value occurs of $ƒ(x,y)$ for $x=\frac{3}{7}$

(D) Minimum value occurs of $ƒ(x,y)$ for $y=\frac{4}{7}$

My approach , all values in the square roots needs to be positive.

${x^2+y^2\ge 0}$ hence it encloses the whole graph

${x^2+y^2-2x+1}$,${(x-1)^2+y^2\ge 0}$

$x^2+y^2-2y+1$, $ x^2+(y-1)^2\ge 0$

$x^2+y^2-6x-8y+25$, $(x-3)^2+(y-4)^2\ge 0$

Not able to approach from here

Samar Imam Zaidi
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1 Answers1

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Hint. Note that the centers of the four circles are: $$O(0,0),A(0,1),B(1,0),C(3,4)$$ For the shortest distance, the point $(x,y)$ must lie on the intersection of the line passing through $(0,0)$ and $(3,4)$ and the line passing through $(0,1)$ and $(1,0)$. Hence: $$|OC|+|AB|=5+\sqrt2.$$

mathcounterexamples.net
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farruhota
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  • Interesting. Can you justify the point $(x,y)$ must lie on the intersection of the line passing through $(0,0)$ and $(3,4)$ and ... – mathcounterexamples.net Oct 18 '19 at 06:38
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    @mathcounterexamples.net Use the Triangle Inequality. It is known in general that, if $ABCD$ is a convex quadrilateral and $P$ is an arbitrary point on the plane, then $$PA+PB+PC+PD\geq AC+BD,.$$ The equality occurs if and only if $P=O$, where $O$ is the intersection of the diagonals $AC$ and $BD$. – Batominovski Oct 18 '19 at 06:47