1
$\begingroup$

Question: Consider a consumer with utility function $U(x,y,z)=y\min\{x,z\}$. The prices of all three goods are the same. The consumer has $100 to spend on these three goods.The demands will be such that:

(a) $y<x=z$

(b) $y>x=z$

(c) $x=y=z$

(d) None of the above

My attempt: The consumer will consume equal amounts of $x$ and $z$ because otherwise the allocation would be inefficient, that is, he can obtain the same level of utility by spending less. So $x=z$. I cannot figure out how is $y$ related to $x$ and $z$. I think the answer would be (d) None of the above because it does not matter if $y$ is less than or greater than or equal to $x$ and $z$.

$\endgroup$
1
  • $\begingroup$ $x=z$ is correct. Now suppose the price of each good is \$1 and the consumer has spent \$99 buying 33 units of each good. She has \$1 left—how should she spend this last dollar? $\endgroup$
    – user18
    Commented Jul 13, 2020 at 4:40

3 Answers 3

2
$\begingroup$

Let $\min\{x,z\}=\Omega$, where $P_\Omega=P_x+P_z$. Now the problem becomes $U(y,\Omega)=y\Omega$, which is a standard Cobb-Douglas with degree 2 of homogeneity. Now in this case the choice for each good is:

$y^*=\frac{\alpha_y100}{P_y(\alpha_y+\alpha_\Omega)}\implies y^*=\frac{100}{2P_y}\;\;\;\;\;\;\;$in this case $\alpha_y=\alpha_\Omega=1$

For $\Omega$: $\;\;\;\;\;\Omega^*=\frac{\alpha_\Omega100}{P_\Omega(\alpha_y+\alpha_\Omega)}\implies \Omega^*=\frac{100}{2P_\Omega} \implies \Omega^*=\frac{100}{2(P_x+P_z)}$

Now, since $P_x=P_y=P_z$, let $P_x=P_y=P_z=P$ a general price, therefore substituting in our optimums:

$y^*=\frac{100}{2P}\;\;\;\;\;\Omega^*=\frac{100}{4P}$

Now it's straightforward (since we already know $x^*=z^*$ and as this is the optimum for $\min\{x,z\}$ which is $x$ OR $z$) that $y^*=\frac{100}{2P}>\Omega^*=\frac{100}{4P}$, so this implies that:

$y^*>x^*=z^*\;\;$****

Also I found this document, where this question is number 13.

Hope this helps.

Disclaimer: It would be helpful if other people could assess this approximation, since I hadn't never seen this problem before.

$\endgroup$
5
  • $\begingroup$ So in the end your solution suggests that x>y for perfectly substitute goods x and y? $\endgroup$
    – user28226
    Commented Jul 12, 2020 at 21:05
  • $\begingroup$ Sorry, I made a little mistake with notation (I assumed it was 𝑈=𝑥min{𝑦,𝑧}), but the conclusion is the same. I will correct those notation mistakes if it makes things more clear, please let me know. The answer should be $y>x=z$ $\endgroup$
    – manifold
    Commented Jul 12, 2020 at 22:37
  • $\begingroup$ of course, this is a game changer! thank you $\endgroup$
    – user28226
    Commented Jul 12, 2020 at 22:41
  • $\begingroup$ I think "$U(x,\Omega)=y\Omega$" should be "$U(y,\Omega)=y\Omega$." $\endgroup$
    – MrAP
    Commented Jul 14, 2020 at 18:16
  • $\begingroup$ @MrAP that's correct! I'll make the changes. $\endgroup$
    – manifold
    Commented Jul 14, 2020 at 18:17
2
$\begingroup$

Hint: Suppose the price of the goods is $P$ so that $N=100/P$ goods can be afforded in total. Now consider which of the following yields more utility:

a) $x=y=z=N/3$.

b) $x=z=N/4$ and $y=N/2$.

$\endgroup$
1
$\begingroup$

I have not seen this in any textbook of mine, but here's my attempt:

Since the utility function (1) is the product of the quantity of y and the minimum quantity of either x or z (so, $min \{x,z\}$ is a singular value, say 15 units or 27 units, etc.) and (2) $x=z$ for every value of x or z, the utility function turns into:

  1. $U(y, x=z)=yx$, or
  2. $U(y, z=x)=yz$

Taking the first case (and the same works for the second), maximization gives the solution for $$ MRS_{XY} = \frac{p_X + p_Z}{p_Y} $$ where $ MRS = dy/dx = MU_x/MU_y = {y}/{x}$, and since $p_X = p_Z = p_Y$ we get $$ y/x = 2 $$ so in the end: $$ y = 2x =2z$$ and overall $y>x=z$.

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

Not the answer you're looking for? Browse other questions tagged or ask your own question.