## anonymous 4 years ago Optimization. Find the minimum distance of a point on the graph xy^2=16 from the origin

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).