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):


$$ U(x,y)=y+100ln(x) $$ $$ M=600 , p_{x}=10 , p_{y}=5 $$

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)


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.


1 Answer 1


Your steps look okay, and the solutions are correct. You can rule out corner solutions by evaluating the utility function at the "corners" and compare those values to the optimum you found. You should be able to verify that $U(50,20)>\max\{U(0,120),\,U(60,0)\}$.

  • $\begingroup$ Is it possible that for U(0,120) there is not utility at all since I would get "120+100ln(0)", which is undefined? If so, how could we interpret that? $\endgroup$ Commented Feb 22, 2021 at 11:34
  • 1
    $\begingroup$ @BachirMessaouri: You're welcome. Yes $U(0,120)$ is undefined. Also note that the MU at 0 is infinity, and so you can use this to rule out the corner where $x =0$. $\endgroup$
    – Herr K.
    Commented Feb 22, 2021 at 12:24

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