# Relation between demands of $x, y$ and $z$

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) $$y (b) $$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$$. •$x=z$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? – Kenny LJ Jul 13 at 4:40

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.

• So in the end your solution suggests that x>y for perfectly substitute goods x and y? – the_rainbox Jul 12 at 21:05
• 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 $y>x=z$ – nrivera Jul 12 at 22:37
• of course, this is a game changer! thank you – the_rainbox Jul 12 at 22:41
• I think "$U(x,\Omega)=y\Omega$" should be "$U(y,\Omega)=y\Omega$." – MrAP Jul 14 at 18:16
• @MrAP that's correct! I'll make the changes. – nrivera Jul 14 at 18:17

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 $$y=N/2$$.

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 $$y>x=z$$.