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  • 4 years ago

Optimization. Find the minimum distance of a point on the graph xy^2=16 from the origin

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  1. anonymous
    • 4 years ago
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    This is your constraint (rewritten): \[f(x)=\sqrt{\frac{16}{x}}\]Your objective function (the value you are trying to minimize) is the distance between some point on the graph and (0,0):\[D=\sqrt{(x_2-x_1)^2+(y_2-y_1)^2}=\sqrt{(0-x)^2+(0-\sqrt{\frac{16}{x}})^2}=\sqrt{(-x)^2+\frac{16}{x}}\]It's enough to minimize the value under the radical, and since it makes the problem a bit simpler, let's go ahead make your O.F all the stuff underneath the radical: \[D=x^2+\frac{16}{x}\]This is the function we are trying to minimize, so let's go ahead and differentiate it: \[\frac{dD}{dx}=2x-\frac{16}{x^2}\]Now we need to find the critical numbers.\[0=2x-\frac{16}{x^2}\rightarrow 2x=\frac{16}{x^2} \rightarrow 2x^3=16 \rightarrow x^3=8 \rightarrow x=2\]This means that the point that is closest to (0,0) is (x,f(x))\[f(x)=\sqrt{\frac{16}{x}} \rightarrow f(2)=\sqrt{\frac{16}{2}}=\sqrt{8}\approx 2.8284\]The point closest to (0,0) in the function xy^2=16 is (2,2.8284).

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