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Find the point on the hyperbola xy=10 closest to the point (3,0). Is there a way to do this without looking at a graph?
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HI!!
probably calculus
I hope so, it's for my Calculus 1A class..
lol then that is what you can use
a point on the graph will look like \((x,\frac{10}{x})\)
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and the square of the distance between that point and \((3,0)\) will be \[\large (x-3)^2+\left(\frac{10}{x}\right)^2\] minimize that in the usual calc way
I know I should take a derivative right?
implicitly or otherwise with respect to x?
If I take a derivative and solve it for zero I get 2x^4 - 6x^3 = 0 .... hmm..
I got x=(1/3)^(5/4) but that can't be right.. I'm not sure how to do this atm..
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