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Calculus/Applied Optimization

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Expert: - 4/24/2010


Question
Hello there, well i got some struggles doing this problem. It's about Max and Min..
Find the Point P on the parabola y=x^2 closest to the point (3,0)

Answer
The distance² from (x,y) to (3,0) is (x-3)² + y².
That is, D² = (x-3)² + y².

If we went to minimize D, then that minimizes D².

Since we know that y = x², then D² is D²(x) = (x-3)² + (x²)².

It can be seen that dD²/dx = 2(x-3) + 2(x²)2x (via the chain rule).
Since the equation is 2x - 6 + 4x³, the answer that jumps out is x = 1 since 2 - 6 + 4 = 0.
Dividing by (x-1) gives (x-1)(4x²+4x+6) = 2(x-1)(2x²+2x+3).
Since b²-4ac is negative on 2x²+3x+3, there are no more solutions.

Thinking about the picture, this must be where a minimum occurs and not a maximum.
Also think about the picture, there is no maximum.

So the closest point on the curve is at x=1, and that is at y=1²=1.  

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