How Do You Solve a Lagrangian Utility Maximization Problem?

[Trizz]

Homework Statement

Find the equations for the utility maximizing values for x and y

U(x,y) = x^2 + y^2

Homework Equations

Budget constraint: I = PxX +Pyy

L(x,y,\lambda ) x^2 + y^2 + \lambda (I - PxX - PyY)

The Attempt at a Solution

I got the three partial derivatives and set equal to zero:

dL/dx = 2x - \lambda Px = 0
dL/dy = 2y - \lambda Py = 0
dL/d\lambda = I-PxX-PyY = 0

Then i set the first two equal to each other to try and find x in terms of y

2x = \lambda Px
2y \lambda Py

This results in x = PxY/PyBut here's the problem...

When I plug that into the last equation, i get stuck

I get:

I - PxX - Py(PyX/Px) = 0

I don't know how to proceed from here algebraically. Normally I'd be able to cancel on some of the simpler problems. But I can't cancel the Px out from the denominatorAny help would be greatly appreciated!

 

[vela]

Remember that $p_x$ and $p_y$ are just constants, so you have an equation of the form $I - a x - b x = 0$ where $a = p_x$ and $b = p_y^2/p_x$. The simplest thing to do would be to multiply through by $p_x$.

 

[Ray Vickson]

Homework Statement

Find the equations for the utility maximizing values for x and y

U(x,y) = x^2 + y^2

Homework Equations

Budget constraint: I = PxX +Pyy

L(x,y,\lambda ) x^2 + y^2 + \lambda (I - PxX - PyY)

The Attempt at a Solution

I got the three partial derivatives and set equal to zero:

dL/dx = 2x - \lambda Px = 0
dL/dy = 2y - \lambda Py = 0
dL/d\lambda = I-PxX-PyY = 0

Then i set the first two equal to each other to try and find x in terms of y

2x = \lambda Px
2y \lambda Py

This results in x = PxY/PyBut here's the problem...

When I plug that into the last equation, i get stuck

I get:

I - PxX - Py(PyX/Px) = 0

I don't know how to proceed from here algebraically. Normally I'd be able to cancel on some of the simpler problems. But I can't cancel the Px out from the denominatorAny help would be greatly appreciated!

The Lagrangian conditions give you x = (λ/2) a and y = (λ/2) b, where I use a and b in place of p_x and p_y. Using these in your constraint gives you everything you need.

That is essentially the way in which most Lagrange multiplier problems are solved, although on rare occasions it is easier to use another method.

RGV