Hi freinds;

can someone give me the solution of #36 on GRE 8767?

I know how I should solve it but I cannot get the answer.

thank you

## GRE 8767# 36

### Re: GRE 8767# 36

We're trying to minimize the function $$d(x,y)=\sqrt{x^2+y^2}$$, subject to the restriction $$xy=8$$, which should immediately suggest the Lagrange multipliers method.

Of course, it's easier to work with $$D(x,y)=x^2+y^2$$. We can minimize this function, then take a square root. Then, letting $$g(x,y)=xy$$, we set $$\nabla D = \langle 2x,2y \rangle = \lambda\nabla g = \lambda\langle x,y \rangle$$, so we have $$2x &= \lambda y,~~~~

2y &= \lambda x,~~~~

8 &= xy.$$

We can quickly work out that the first two equations are satisfied when either $$y=x$$ or $$y=-x$$. But $$y=-x$$ can never satisfy $$8=xy$$, so we must have $$y=x$$, and thus $$8=x^2$$ and $$8=y^2$$, so $$D(x,y)=16$$, and $$d(x,y)=4$$.

This is the "this is a timed test and we're in a hurry" version. I'm sure someone else could give a more lucid explanation.

Of course, it's easier to work with $$D(x,y)=x^2+y^2$$. We can minimize this function, then take a square root. Then, letting $$g(x,y)=xy$$, we set $$\nabla D = \langle 2x,2y \rangle = \lambda\nabla g = \lambda\langle x,y \rangle$$, so we have $$2x &= \lambda y,~~~~

2y &= \lambda x,~~~~

8 &= xy.$$

We can quickly work out that the first two equations are satisfied when either $$y=x$$ or $$y=-x$$. But $$y=-x$$ can never satisfy $$8=xy$$, so we must have $$y=x$$, and thus $$8=x^2$$ and $$8=y^2$$, so $$D(x,y)=16$$, and $$d(x,y)=4$$.

This is the "this is a timed test and we're in a hurry" version. I'm sure someone else could give a more lucid explanation.