# Thread: How to get the solution of this simple optimization problem?

1. ## How to get the solution of this simple optimization problem?

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

2. 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.

3. 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