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## Homework Statement

Find the point(s) on the surface z

^{2}- 10xy = 10 nearest to the origin.

## The Attempt at a Solution

I will minimize distance by minimizing distance

^{2}. So,

D = x

^{2}+ y

^{2}+ z

^{2}, where z

^{2}- 10xy = 10. By Lagrange multipliers I have,

2x = -10yλ

2y = -10xλ

2z = 2zλ.

By the last equation, clearly either z = 0 or λ = 1 or λ = 0.

I will test λ = 1 first with the condition y ≠ 0 and x ≠ 0. By dividing the first equation by the second equation I have,

x/y = y/x, which implies x = y. But clearly 2x ≠ -10x when x ≠ 0. So I reject λ = 1 with the condition y ≠ 0 and x ≠ 0. I will now test λ = 1 first with the condition y = 0. I have,

2(0) = 10 * x * 1, so clearly x = 0. (Similarly, I could have tested with the condition x = 0 and found y = 0.) By the constraint equation I have z

^{2}- 10(0)(0) = 10, so clearly z = ±√10.

Suppose z = 0. Then x ≠ 0 and y ≠ 0 because, in our constraint equation, -10xy = 10. So clearly xy = -1. I will now divide 2x = -10yλ by 2y = -10xλ and again find x = y. This implies,

x

^{2}= -1 or y

^{2}= -1.

x = i or y = i. I reject these solutions as crazy.

Suppose λ = 0. Then it must follow that y = 0 and x = 0. Again, plugging into our constraint equation gives z = ±√10. Therefore, Lagrange multipliers give the points (0,0,±√10), both of which give distance √10.

However, by inspection, one can see that the points (1,-1,0) and (-1,1,0) also satisfy the constraint and give distance √2.

Ideas where I went wrong?