Find The Minimum!!!!!!!

Geometry Level 4

A Straight Line L With Negative Slope Passes Through The Point (8,2) And Cuts The Positive Coordinate Axes At Points P And Q. Find The Absolute Minimum Value Of OP+OQ ,as L Varies where O Is The Origin. Absolute Value Means Magnitude for example absolute value of -2 is 2. This Problem Is An Old IITJEE Problem.


The answer is 18.

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Prakhar Bindal by Prakhar Bindal , 22, India

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2 solutions

Aryan Goyat
Nov 14, 2015

the equation of line cutting x axis at (a,0) and y axis at (0,b): x/a+y/b=1 since line passes through(8,2)---8/a+2/b=1 ------to find minimum of a+b ------ using Engel's inequality 8/a+2/b>={root(8)+root(2)}*{root(8)+root(2)}/a+b ---mini=18

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Abhishek Sharma
Nov 26, 2014

Assume the slope of line to be m such that m<0.

Now the equation of the line using point slope method is, mx-y=8m-2.

The coordinates of P and Q respectively are ( 8 − 2 m , 0 ) (8-\frac { 2 }{ m } ,0) and ( 0 , 2 − 8 m ) (0,2-8m) .

O P + O Q = − 8 m 2 + 10 m − 2 m OP+OQ=\frac { -8{ m }^{ 2 }+10m-2 }{ m } .

Taking its derivative and putting it equal to zero we get m=-1/2 and m=1/2.

As m<0, m=-1/2.

Now put m=-1/2 in the expression to get minimum value of OP+OQ 18.

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Exactly The Same Method I Solved It When I Saw this for the same time.but instead of taking derivative i used AM GM Concept.

Prakhar Bindal - 6 years, 6 months ago

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