# Calc-optimization

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Given y=(x)^1/2, find the closest point to (3/2,0)

• Calc-optimization -

let 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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