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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 $z$.

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  • $\begingroup$ $x=z$ 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? $\endgroup$ – Kenny LJ Jul 13 at 4:40
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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.

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  • $\begingroup$ So in the end your solution suggests that x>y for perfectly substitute goods x and y? $\endgroup$ – the_rainbox Jul 12 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 $y>x=z$ $\endgroup$ – nrivera Jul 12 at 22:37
  • $\begingroup$ of course, this is a game changer! thank you $\endgroup$ – the_rainbox Jul 12 at 22:41
  • $\begingroup$ I think "$U(x,\Omega)=y\Omega$" should be "$U(y,\Omega)=y\Omega$." $\endgroup$ – MrAP Jul 14 at 18:16
  • $\begingroup$ @MrAP that's correct! I'll make the changes. $\endgroup$ – nrivera Jul 14 at 18:17
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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 $y=N/2$.

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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 $y>x=z$.

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