Minimum value.

Geometry Level pending

A straight line L L with negative slope passes through the point K ( 8 , 2 ) K(8,2) and cuts the positive coordinate axes at P P and Q Q . Find the absolute minimum value of O P + O Q OP+OQ , as L L varies.

Details and assumptions:

-Take O O as the origin.

- P P and Q Q cuts the x x -axis and y y -axis, respectively.


The answer is 18.

Sahil Silare by Sahil Silare , 20, India

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1 solution

Saya Suka
Dec 16, 2016

Let the distance between P and (8,0) be x and distance between Q and (0,2) be y.
y/8 = 2/x
y = 16/x

OP + OQ
= (8+x) + (2+y)
= 10 + x + y
= 10 + x + 16/x
(OP + OQ)'
= 1 - 16/x^2 = 0
x = 4 = y
OP + OQ
= 10 + 4 + 4 = 18

2 Helpful 0 Interesting 0 Brilliant 0 Confused

Can you send figure

Divyanshu Sahu - 4 years, 6 months ago

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Asking for solution is wrong @Divyanshu Sahu . Please review it @Calvin Lin

Sahil Silare - 4 years, 6 months ago

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He's asking you to clarify what you're doing, ideally with a figure of the setup.

I agree that right now, your solution is hard to understand at a glance, and adding an image would be immensely helpful.

Calvin Lin Staff - 4 years, 5 months ago

Nice solution @Saya Suka but it would be more pretty if you have used LaTeX.

Sahil Silare - 4 years, 6 months ago

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