1. ## optimization (multivariable) Done 2 ways, but diff answers

"Find the point on the paraboloid :

z=(x^2)/25 + (y^2)/4

That is closest to the point (0,5,0)"

...my solutions manual solves this using symetry. they say that by symetry we see that x=0. And then the problem turns into a two dimensional problem. and optimize the distance squared... they got (0,2.66,1.76) as the closest point.

...NOW, i did this in 3D using the paraboloid as my constraint, and did it by the method of lagrange multipliers. but i get a totally diff answer were some of my points are negative.

2. You say you used Langrange multipliers but you don't show how you did that. All I can say is that, using Lagrange multipliers, I get (0, 5, 0).

3. I really hope i did some dumb mistake bc this problem is frustating me.

Ok, my function to be optimized is the square of the distance formula, between the (0,5,0) and a point x,y,z on the paraboloid: d^2= F = x^2 + (y-5)^2 + z^2 and my constraint will be the parabloid function set equal to zero: (x^2)/25 + (y^2)/4 - z = 0. and then i did the partial integrations, setting them equal to the partials of the constraint times lambda. Is there something wrong in my analysis?

4. Originally Posted by HallsofIvy
You say you used Langrange multipliers but you don't show how you did that. All I can say is that, using Lagrange multipliers, I get (0, 5, 0).
And i'm not sure (0,5,0) makes sense bc that point isnt in the paraboloid

5. Your are right. I don't know where my head was! (0, 5, 0) was the point we are measuring the distance to, not the answer.

The problem is to minimize $f(x,y,z)= x^2+ (y- 5)^2+ z^2$ subject to the constraint $g(x, y, z)= \frac{x^2}{25}+ \frac{y^2}{4}- z= 0$. $\nabla f= 2x\vec{i}+ 2(y- 5)\vec{j}+ 2z\vec{k}= \lambda\nabla g= \lambda(\frac{2}{25}x\vec{i}+ \frac{1}{4}y\vec{j}- \vec{k}$ or $2x= \lambda \frac{2}{25}x$, $2(y- 5)= \lambda\frac{1}{4}y$, and $2z= -\lambda$.

If you divide the first equation by the last, $\frac{x}{z}= \frac{2}{25}x$ so x= 0 or $z= \frac{2}{25}$. If you divide the first equation by the second, $\frac{x}{y- 5}= \frac{20}{25}\frac{x}{y}$ so $5xy= 4x(y- 5)= 4xy- 20$ so xy= 20.

6. Originally Posted by HallsofIvy
If you divide the first equation by the last, $\frac{x}{z}= \frac{2}{25}x$ so x= 0 or $z= \frac{2}{25}$. If you divide the first equation by the second, $\frac{x}{y- 5}= \frac{20}{25}\frac{x}{y}$ so $5xy= 4x(y- 5)= 4xy- 20$ so xy= 20.
I'm sorry for the persistance but when i divide the equations like u said, i got z=-25/2 and y=-40/17...now the problem is that when i plug this into the constraint, i get no solution when solving for x