Lagrangian multiplier and optimal bundle I would like to know where I am wrong (if I am) and why I am wrong here please:
If a consumer has an income of 600 euros to spend for good x (Px = 10 euros) and good y (Py = 5 euros).
What is the optimal bundle considering that U(x,y) = y + 100 ln(x)?
Here is how I did it (but I'm not confident):
Data:
$$
U(x,y)=y+100ln(x)
$$
$$
M=600 , p_{x}=10 , p_{y}=5
$$
Applied Lagrange equation:
$$ 
\mathcal{L}= U(x,y) - \lambda (p_{x}x+p_{y}y-M)
$$
$$ 
\mathcal{L}= y+100ln(x) - \lambda (10x+5y-600)
$$
Condition 1:
$$\partial U/\partial x-10\lambda = 0$$
$$100/x -10\lambda = 0$$
Condition 2:
$$\partial U/\partial y -5\lambda = 0$$
$$1 -5\lambda = 0$$
$$(\lambda = 1/5)$$
Using a system of equations I get:
$$ 100/x - 10\lambda = 2-10\lambda$$
$$ x  = 50$$
Now, plugging this into Condition 3:
$$M-p_{x}x-p_{y}y=0$$
$$600-500-5y=0$$
$$y=20$$
Therfore xMax = 50 and yMax = 20
The optimal bundle is (50,20)
Comment:
I'm so not sure about this...
I'm skeptical because condition 2 doesn't seem to show the y variable, which seems to indicate that y should be equal to 0 at that point (so we should expect a corner solution).
But then, as I found x after equating the two first conditions and plugged it in the third condition, it gave me y=20 after all.
So I'm kind of confused.
If both lambda in cond1 and cond2 are indeed equal to 1/5, my answer (y=20) should be ok but if it's not the case, it might very well be that y=0.
I don't know how to check my answer to be sure I'm right...or wrong.
And, if by chance, I got it right, can we rule out corner/boundary solutions anyway?
Thank you for your help.
Your steps look okay, and the solutions are correct. You can rule out corner solutions by evaluating the utility function at the &quot;corners&quot; and compare those values to the optimum you found. You should be able to verify that $U(50,20)&gt;\max\{U(0,120),\,U(60,0)\}$.