<p>I think either x or z will be zero when utility is maximized if price of x and z are not equal.
But if Px=Pz then you can consume x=y=z=M/(px+py+pz) or any other combination on the budget line that statisfies either x=y or y=z or both.</p>

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2022-10-17T09:15:11.163Z
2022-10-17T09:15:11.163Z
abc
<p>Think about the problem from the following perspective:</p>
<p>You have 2 <em>bundles</em> you can buy. The first one is <span class="math-container">$\{x,y\}$</span> while the second is <span class="math-container">$\{y,z\}$</span>. It does not make sense to purchase only one element from each bundle since you need to consume both goods in the <em>bundle</em> together to gain at least some utility from it. Goods inside the <em>bundle</em> are <strong>perfect complements</strong> but these two bundles themselves are <strong>perfect substitutes</strong> to one another.</p>
<p>Thus it must hold that if the price of the first <em>bundle</em> is higher than the price of the second one, you have no incentive to pay for the first one. And <em>vice versa</em>.</p>
<p>Due to the good <span class="math-container">$y$</span> being in both <em>bundles</em>, you have to compare only the price of good <span class="math-container">$x$</span> with the price of good <span class="math-container">$z$</span>.</p>

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2022-10-17T19:00:13.507Z
2022-10-17T19:05:59.350Z
2022-10-17T19:05:59.350Z
Athaeneus
<p>My attempt.</p>
<p>I think that cannot be said in advance if <span class="math-container">$x=t=z$</span> is the optimal solution or not, bur it depends on the prices of the goods.</p>
<p>First, notice that can't be <span class="math-container">$x=z=0$</span>, otherwise <span class="math-container">$\max U$</span> will be <span class="math-container">$0$</span>.
And, for the same reason, must  be <span class="math-container">$y\neq0$</span>.</p>
<p>So, we have two cases:</p>
<ol>
<li><p><span class="math-container">$x\neq0$</span>, <span class="math-container">$y\neq0$</span>, <span class="math-container">$z\neq0$</span>, <span class="math-container">$y\neq0$</span>..</p>
</li>
<li><p><span class="math-container">$x$</span> or <span class="math-container">$z$</span> are <span class="math-container">$0$</span> (it's the same), say <span class="math-container">$z=0$</span>, <span class="math-container">$y\neq0$</span>.</p>
</li>
</ol>
<p>Let <span class="math-container">$p_xx+p_yy+p_zz=K$</span> be the budget line , <span class="math-container">$K$</span> is income.</p>
<p><strong>First case</strong>, <span class="math-container">$x\neq0$</span>, <span class="math-container">$y\neq0$</span>, <span class="math-container">$z\neq0$</span>.</p>
<p>In this case the <span class="math-container">$U$</span> is maximized if <span class="math-container">$x=y=z$$^{[1]}$</span>.</p>
<p>From the budget equation, the optimal solution is :</p>
<p><span class="math-container">$x=y=z=\frac{K}{p_x+p_y+p_z}$</span>,</p>
<p>and the corresponding value of the objective function <span class="math-container">$U$</span>, say <span class="math-container">$U_1$</span>, is:</p>
<p><span class="math-container">$U_1=\frac{2K}{p_x+p_y+p_z}$</span>.</p>
<p><strong>Second case.</strong> <span class="math-container">$z=0$</span>, <span class="math-container">$x\neq0, y\neq0$</span>.</p>
<p>In this case the <span class="math-container">$U$</span> is maximized if <span class="math-container">$x=y^{[1]}$</span>.</p>
<p>From the budget equation, the optimal solution is :</p>
<p><span class="math-container">$x=y=\frac{K}{p_x+p_y}$</span>,</p>
<p>and the corrisponding value of the objective function <span class="math-container">$U$</span>, say <span class="math-container">$U_2$</span> is:</p>
<p><span class="math-container">$U_2=\frac{K}{p_x+p_y}$</span>.</p>
<p>Therefore, <span class="math-container">$U_1\leq U_2 \Leftrightarrow $</span>
<span class="math-container">$(\frac{2K}{p_x+p_y+p_z}\leq \frac{K}{p_x+p_y}) \Leftrightarrow (p_z\geq p_x+p_y$</span>).</p>
<p>Therefore, it would be convenient choose <span class="math-container">$z=0$</span> if the last inequality is satisfied. Otherwise, it is convenient the other solution.</p>
<p><span class="math-container">$$***$$</span></p>
<p>From an intuitive point of view, one can say that if the price of a good, in our case <span class="math-container">$z$</span>, is 'very high', it is convenient set <span class="math-container">$z=0$</span> and to concentrate our purchases on the other two goods.</p>
<p><span class="math-container">$----------------------$</span></p>
<p><strong>[1]</strong> It should be justified, but I think it is enough intuitive.</p>

CC BY-SA 4.0
2022-10-17T19:18:01.493Z
2022-10-17T19:18:01.493Z
BakerStreet
<p>Suppose I have 3 goods, x, y, and z and my utility function is given by U(x, y, z) = min(x, y) + min(y, z). At the optimal consumption level, is it accurate to say that x = y = z?</p>
<p>My reasoning here is that since each minimum term is determined by the smallest value, it makes no sense to consume more of the other good. So for the first term, the optimal consumption level of x and y is when x = y, and for the second term the optimal consumption is z = y, but I'm not entirely sure if this is the right way to think about it.</p>

CC BY-SA 4.0
2022-10-18T09:13:24.240Z
The trick is to write down all the cases first and look at the utility function based on each case. For instance if $x > y$ and $y > z$, then $u = y + z$ and $x = 0$. There are 9 cases if you consider but you can reduce it down if you get smart with your inequalities and use $\geq$ and $\leq$
Rumi
<microeconomics>
Optimal consumption of a specific utility function
test4355
# Optimal consumption of a specific utility function

(Asked by: username_0 on 2022-10-17)

Suppose I have 3 goods, x, y, and z and my utility function is given by U(x, y, z) = min(x, y) + min(y, z). At the optimal consumption level, is it accurate to say that x = y = z?

My reasoning here is that since each minimum term is determined by the smallest value, it makes no sense to consume more of the other good. So for the first term, the optimal consumption level of x and y is when x = y, and for the second term the optimal consumption is z = y, but I'm not entirely sure if this is the right way to think about it.

**Tags:** microeconomics
> **Comments:**
> **username_1**: The trick is to write down all the cases first and look at the utility function based on each case. For instance if $x > y$ and $y > z$, then $u = y + z$ and $x = 0$. There are 9 cases if you consider but you can reduce it down if you get smart with your inequalities and use $\geq$ and $\leq$  

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## 3 answers

### Answer by username_2 on 2022-10-17. Score: 1

I think either x or z will be zero when utility is maximized if price of x and z are not equal.
But if Px=Pz then you can consume x=y=z=M/(px+py+pz) or any other combination on the budget line that statisfies either x=y or y=z or both.


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### Answer by username_3 on 2022-10-17. Score: 1

Think about the problem from the following perspective:

You have 2 *bundles* you can buy. The first one is $\{x,y\}$ while the second is $\{y,z\}$. It does not make sense to purchase only one element from each bundle since you need to consume both goods in the *bundle* together to gain at least some utility from it. Goods inside the *bundle* are **perfect complements** but these two bundles themselves are **perfect substitutes** to one another.

Thus it must hold that if the price of the first *bundle* is higher than the price of the second one, you have no incentive to pay for the first one. And *vice versa*.

Due to the good $y$ being in both *bundles*, you have to compare only the price of good $x$ with the price of good $z$.


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### Answer by username_4 on 2022-10-17. Score: 0

My attempt.

I think that cannot be said in advance if $x=t=z$ is the optimal solution or not, bur it depends on the prices of the goods.

First, notice that can't be $x=z=0$, otherwise $\max U$ will be $0$.
And, for the same reason, must be $y\neq0$.

So, we have two cases:

1. $x\neq0$, $y\neq0$, $z\neq0$, $y\neq0$..
2. $x$ or $z$ are $0$ (it's the same), say $z=0$, $y\neq0$.

Let $p\_xx+p\_yy+p\_zz=K$ be the budget line , $K$ is income.

**First case**, $x\neq0$, $y\neq0$, $z\neq0$.

In this case the $U$ is maximized if $x=y=z$$^{[1]}$.

From the budget equation, the optimal solution is :

$x=y=z=\frac{K}{p\_x+p\_y+p\_z}$,

and the corresponding value of the objective function $U$, say $U\_1$, is:

$U\_1=\frac{2K}{p\_x+p\_y+p\_z}$.

**Second case.** $z=0$, $x\neq0, y\neq0$.

In this case the $U$ is maximized if $x=y^{[1]}$.

From the budget equation, the optimal solution is :

$x=y=\frac{K}{p\_x+p\_y}$,

and the corrisponding value of the objective function $U$, say $U\_2$ is:

$U\_2=\frac{K}{p\_x+p\_y}$.

Therefore, $U\_1\leq U\_2 \Leftrightarrow $
$(\frac{2K}{p\_x+p\_y+p\_z}\leq \frac{K}{p\_x+p\_y}) \Leftrightarrow (p\_z\geq p\_x+p\_y$).

Therefore, it would be convenient choose $z=0$ if the last inequality is satisfied. Otherwise, it is convenient the other solution.

$$\*\*\*$$

From an intuitive point of view, one can say that if the price of a good, in our case $z$, is 'very high', it is convenient set $z=0$ and to concentrate our purchases on the other two goods.

$----------------------$

**[1]** It should be justified, but I think it is enough intuitive.


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