How to get the solution of this simple optimization problem?

**ggyyree** · November 8th 2009, 07:40 AM

How to get the solution of this simple optimization problem?

$ minimize~~\Phi = x_1^2 + x_2^2 $
$ subject~to~~x_1 +2x_2 = 5 $

I konw we may use Moore-Penrose inverse like,

$ Ax = b => x = (A^TA)^{-1}A^Tb $

but what are $A$ and $x$ now?

if $A = (1~~2)$ and $A^T = (1~~2)^T$ then there is no inverse of $A^TA$ . I am quite confused how to formulate these $A$ and $x$ to do the matrix multiplication.

Thanks a lot

**HallsofIvy** · November 11th 2009, 02:50 AM

Originally Posted by ggyyree

How to get the solution of this simple optimization problem?

$ minimize~~\Phi = x_1^2 + x_2^2 $
$ subject~to~~x_1 +2x_2 = 5 $

I konw we may use Moore-Penrose inverse like,

$ Ax = b => x = (A^TA)^{-1}A^Tb $

but what are $A$ and $x$ now?

if $A = (1~~2)$ and $A^T = (1~~2)^T$ then there is no inverse of $A^TA$ . I am quite confused how to formulate these $A$ and $x$ to do the matrix multiplication.

Thanks a lot

"Moore-Penrose" is much too advanced for me ("Moore" isn't so bad but I cringe when I hear "Penrose") so I would do it with Lagrange multipliers (I much better with "Lagrange").

You want to minimize $F(x,y)= x^2+ y^2$ subject to the constraint G(x,y)= x+ 2y= 5. The gradients are $\nabla F= 2x\vec{i}+ 2y\vec{j}$ and $\nabla G= \vec{i}+ 2\vec{j}$ . Now we need to find (x,y) so that $2x\vec{i}+ 2y\vec{j}= \lambda(\vec{i}+ 2\vec{j})$ . That means we must have $2x= \lambda$ and $2y= 2\lambda$ . Dividing the second equation by the first give $\frac{y}{x}= 2$ or y= 2x. Putting that into the constraint x+ 2y= 5, x+ 2(2y)= x+ 4x= 5x= 5 or x= 1, y= 2.

**CaptainBlack** · November 11th 2009, 04:47 AM

Originally Posted by ggyyree

How to get the solution of this simple optimization problem?

$ minimize~~\Phi = x_1^2 + x_2^2 $
$ subject~to~~x_1 +2x_2 = 5 $

Why not solve the constraint for $x_1$ in terms of $x_2$ and substitute into the objective to reduce this to a 1-D optimisation??

$x_1=5-2x_2$ , then:

$ \Phi=5x_2^2-20x_2+25 $

CB

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