SOLUTION: **Solve using calculus optimization** Find the point on the curve {{{ y=x^2 }}} that is closest to the point (18, 0).

Algebra ->  Equations -> SOLUTION: **Solve using calculus optimization** Find the point on the curve {{{ y=x^2 }}} that is closest to the point (18, 0).      Log On


   



Question 1142727: **Solve using calculus optimization**
Find the point on the curve +y=x%5E2+ that is closest to the point (18, 0).

Answer by Alan3354(69443) About Me  (Show Source):
You can put this solution on YOUR website!
Find the point on the curve +y=x%5E2+ that is closest to the point (18, 0).
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The points on the curve are (x,x^2)
The distance d(x) to (18,0) is sqrt%28%28x-18%29%5E2+%2B+%28x%5E2-0%29%5E2%29
d(x) = sqrt%28x%5E4+%2B+x%5E2+-+36x+%2B324%29
d'(x) = (1/2)*(4x^3 + 2x - 36)/sqrt(...) = 0
2x^3 + x - 18 = 0
x = 2
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The point is (2,4)
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The distance is sqrt(16^2 + 4^2) = 4sqrt(17) units.