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January 28, 2015

January 28, 2015

Posted by **Daryl** on Saturday, December 15, 2012 at 9:28pm.

- Calc-optimization -
**Reiny**, Saturday, December 15, 2012 at 11:39pmlet the closest point be P(a,b)

label the given point A(3/2 , 0)

AP will have a minimum length when AP is perpendicular to the tangent at P

slope of tangent = y' = (1/2)x^(-1/2) or 1/(2√x)

so at P the slope of the tangent is 1/2√a

slope of AP = b/(a-3/2)

b/(a-3/2) = -2√a , (the negative reciprocal of 1/2√a

but also b = √a

√a/(a-3/2) = -2√a

1/(a-3/2) = -2

1 = -2a + 3

2a = 2

a = 1

then b=√1 = 1

the closest point is (1,1)

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