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October 31, 2014

October 31, 2014

Posted by **Sara** on Wednesday, October 30, 2013 at 12:42pm.

Note: Your answer should be a point in the xy-plane, and as such will be of the form (x-coordinate,y-coordinate)

- Calculus-Applied Optimization Problem: -
**Steve**, Wednesday, October 30, 2013 at 1:39pmwhat a stupid note. Of course it will be a point with (x,y) coordinates! This is a calculus class - you know all that already.

Given a point (x,y) on the line, y=1-2x

The distance from (x,1-2x) to (3,1) is

d^2 = (x-3)^2 + (1-(1-2x))^2 = 5x^2-6x+9

so, we want minimum d, when dd/dx = 0:

2d dd/dx = 10x-6

dd/dx = 10x-6/2sqrt(blah blah)

dd/dx=0 when x = 3/5.

So, minimum d^2 is

5(3/5)^2 - 6(3/5) + 9 = 36/5

minimum d is 6/√5

Or, as we all know, the distance from a point (x,y) to a line ax+by+c=0 is

|ax+by+c|/√(a^2+b^2) = (6(3)+3(1)-3)/√(36+9) = 18/√45 = 6/√5

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