calculus
posted by Anonymous .
Find the point on the line 4x+2y+5=0 which is closest to the point (2,1)

calculus 
Steve
without calculus, the distance from the point to the line is
4(2)+2(1)+5/√(4^2+2^2) = √5/2
Using calculus, we have the distance z from (2,1) to (x,y) is
z^2 = (x+2)^2 + (y+1)^2
But y = (4x+5)/2, so
z^2 = (x+2)^2 + (((4x+5)/2)+1)^2
z^2 = 5x^2 + 10x + 25/4
z = √(5x^2 + 10x + 25/4)
dz/dx = 10(x+1)/√(5x^2+10x+25/4)
dz/dx=0 at x=1
z(1) = √5/2
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