A line through the point (2,2) cuts the x and y axes at A and B. Find the minimum length of the segment AB.

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- Apr 13th 2010, 12:19 PMrobertosavinOptimization Segment Minimum
A line through the point (2,2) cuts the x and y axes at A and B. Find the minimum length of the segment AB.

- Apr 13th 2010, 03:07 PMzzzoak
OA=2+x x-distance from x=2 to the right

tg a = 2/x

OB=(2+x)tg a=2(2+x)/x

$\displaystyle |AB|^2=|OA|^2+|OB|^2$

f(x)=$\displaystyle |AB|^2=(2+x)^2+\frac{4(2+x)^2}{x^2}$

f'(x)=0

$\displaystyle x^3-8=0$

$\displaystyle x=2$

OA=4 (x)

OB=4 (y).

|AB| min=$\displaystyle 4\sqrt{2}$ - Apr 13th 2010, 03:28 PMArchie Meade
The line has a negative slope.

If it had a positive slope, it could go through the origin making the distance

between A and B zero.

If you draw the line going through (2,2) with a negative slope,

we have similar triangles by drawing lines from (2,2) to the x and y axes.

hence $\displaystyle tan\theta=\frac{2}{x-2}=\frac{y-2}{2}$

$\displaystyle y-2=\frac{4}{x-2}$

let x-2=c, y-2=k

the length of the line segment is, using Pythagoras' theorem

$\displaystyle \sqrt{(k)^2+2^2}+\sqrt{(c)^2+2^2}=\sqrt{\frac{4^2} {c^2}+4}+\sqrt{c^2+4}$

You could differentiate this with respect to c and set the result = 0

to find the value of x causing the segment length to be a minimum.

Alternatively,

the segment length is

$\displaystyle \frac{2}{cos\theta}+\frac{2}{sin\theta}$

we can differentiate wrt the angle to find the minimum length

$\displaystyle \frac{2sin\theta}{cos^2\theta}-\frac{2cos\theta}{sin^2\theta}=0$

$\displaystyle \frac{2sin^3\theta-2cos^3\theta}{sin^2\theta\ cos^2\theta}=0$

$\displaystyle sin^3\theta-cos^3\theta=0$

this occurs when the angle is 45 degrees

hence the minimum segment length is $\displaystyle 2sin45^o+2cos45^o$

or $\displaystyle \sqrt{4^2+4^2}$