<p>I assume <span class="math-container">$x_i$</span> represents the quantity and belongs to <span class="math-container">$\mathbb R_{+}$</span>. You can form the constraints as follows:
<span class="math-container">$$
x_i \geq0 \quad\forall i \in [3] \\
\sum_{i=1}^3p_ix_i \leq I
$$</span>
You can simplify the objective by noting that for the utility to be the maximum, <span class="math-container">$x_2 =x_3$</span>. Try to reason why this is true. Hence, the final problem becomes,</p>

<p><span class="math-container">$$\
\max_{x_1, x_2, x_3}x_1^{1/3} +x_2 \quad s.t. \\
x_i \geq0 \quad\forall i \in [3] \\
\sum_{i=1}^3p_ix_i \leq I \\
x_2=x_3
$$</span>
Can you now set up the lagrange multiplier function? As far as KKT conditions are concerned, I believe they should help because you are trying to maximize a concave function(in <span class="math-container">$\mathbb R_+$</span>) subject to convex constraints and the strict inequality constraint is linear.</p>

<p><strong>Edit 1:</strong> Let's eliminate <span class="math-container">$x_3$</span> as we know that <span class="math-container">$x_2=x_3$</span>. The problem simplifies to
<span class="math-container">$$\
\min_{x_1,\ x_2}\ -x_1^{1/3} -x_2 \quad s.t. \\
x_1 \geq0,\ x_2 \geq 0 \\
p_1x_1 + (p_2 + p_3)x_2 \leq I
$$</span>
<span class="math-container">$$\
\mathcal L(x_1, x_2)=-x_1^{1/3} -x_2 + \lambda_1(-x_1) + \lambda_2(-x_2) + \lambda(p_1x_1 + (p_2 + p_3)x_2 - I)
$$</span>
<strong>Edit 2:</strong> I see solving the Lagrangian can be tough. Let's simplify, we know that all constraints cannot be active at the same time. It will yield an utility of zero.</p>

<p><strong>Case-1:</strong> <span class="math-container">$x_1=0, x_2&gt;0$</span>
You can check algebraically, <span class="math-container">$x_2=\frac{I}{p_2+p_3}$</span>. Hence, utility<span class="math-container">$(U)=\frac{I}{p_2+p_3}$</span></p>

<p><strong>Case-2:</strong> <span class="math-container">$x_1&gt;0, x_2=0$</span>
You can check algebraically, <span class="math-container">$x_1=\frac{I}{p_1}$</span>. Hence, utility<span class="math-container">$(U)=\left(\frac{I}{p_1}\right)^{1/3}$</span></p>

<p><strong>Case-3</strong> <span class="math-container">$x_1&gt;0, x_2&gt;0$</span>
You can now use the Lagrangian, its much simpler than the original Lagrangian problem because <span class="math-container">$\lambda_1=\lambda_2=0$</span>
<span class="math-container">$$\
\mathcal L(x_1, x_2)=-x_1^{1/3} -x_2 + \lambda(p_1x_1 + (p_2 + p_3)x_2 - I)\\
\frac{\partial\mathcal L}{\partial x_1} = 0 \implies x_1 = \left(\frac{1}{3\lambda p_1}\right)^{3/2}\\
\frac{\partial\mathcal L}{\partial x_2} = 0 \implies \lambda = \frac{1}{p_2+p_3}\\
\text{Use the constraint, }p_1x_1 + (p_2 + p_3)x_2 = I \text{ to find } x_2
$$</span>
You can simply finish saying the solution is <span class="math-container">$\max$</span> of all three cases. As per the KKT conditions are concerned, one of the three solutions will satisfy KKT Conditions depending on <span class="math-container">$p_1, p_2, p_3 \text{ and } I$</span>. Note that this is no different from the original problem and hence, KKT condition should be satisfied by one of the three depending on <span class="math-container">$p_1, p_2, p_3 \text{ and } I$</span>. We could have obtained the same solution using the original Lagrangian. Even there we would have ended up with a <span class="math-container">$\max$</span> because of <span class="math-container">$\lambda_1, \lambda_2$</span> and invoking KKT.</p>

CC BY-SA 4.0
2020-06-14T10:15:21.813Z
2020-06-14T11:02:14.543Z
2020-06-14T11:02:14.543Z
Shiv Tavker
<p>I try to use substitution instead of relying on KKT or the Lagrangian method until I encounter a problem that is better solved through them. In a sense, it is like people preferring different methods like elimination or substitution for solving a simple system of equations, but here we are talking about optimization problems instead of systems of equations</p>
<p>We can solve the problem you provided using substitution and simplifying the optimization problem.</p>
<p>Our UMP in consideration is:
<span class="math-container">$$\begin{aligned}
&amp;\max_{x_1,x_2,x_3\geq0} \quad  x_1^\frac{1}{3} + \min(x_2,x_3)\\
&amp;\textrm{s.t.} \quad  p_1x_1+p_2x_2+p_3x_3\leq I \end{aligned}$$</span></p>
<p>Now, what I can do is first solve the UMP holding <span class="math-container">$x_1$</span> constant to get demands for <span class="math-container">$x_2, x_3$</span> as functions of <span class="math-container">$x_1$</span> and other parameters:
<span class="math-container">$$\begin{aligned}
&amp;  \max_{x_1}  {\begin{aligned} \begin{pmatrix} \displaystyle\max_{x_2,x_3} &amp; x_1^\frac{1}{3}+\min(x_2,x_3) \\ \textrm{s.t.}  &amp; p_2x_2+p_3x_3\leq I-p_1x_1 \end{pmatrix} \end{aligned}}\\ \end{aligned}$$</span>
Notice that UMP involving <span class="math-container">$x_2,x_3$</span> is a standard problem over perfect complements. The problem inside the brackets gives us:
<span class="math-container">$$x_2(x_1,p,I)=x_3(x_1,p,I)=\frac{I-p_1x_1}{p_2+p_3}$$</span></p>
<p>We can now use the above, substitute them into the problem inside the bracket of the original UMP and solve the optimization to get optimal <span class="math-container">$x_1^*$</span>:
<span class="math-container">$$\begin{aligned}
&amp;\max_{0 \leq x_1\leq\frac{I}{p_1}} \quad  x_1^\frac{1}{3}+\frac{I-p_1x_1}{p_2+p_3}\\
\text{gives} \quad &amp; x_1^*(p,I)=\left({\frac{p_2+p_3}{3p_1}}\right)^\frac{3}{2}
\end{aligned}$$</span></p>
<p>Substituting <span class="math-container">$x_1^*$</span> in the expression for <span class="math-container">$x_2,x_3$</span> that we derived before gives us the final demands:</p>
<p><span class="math-container">$$(x_1^*,x_2^*,x_3^*)(p,I)=\begin{cases}
\left (\left({\frac{p_2+p_3}{3p_1}}\right)^\frac{3}{2},\frac{I}{p_2+p_3}-\frac{1}{3}\left(\frac{p_2+p_3}{3p_1}\right)^\frac{1}{2}, \frac{I}{p_2+p_3}-\frac{1}{3}\left(\frac{p_2+p_3}{3p_1}\right)^\frac{1}{2}\right) &amp; \text{if } I\sqrt{p_1} &gt; \left(\frac{p_2+p_3}{3}\right)^\frac{3}{2}\\
\left (\left({\frac{p_2+p_3}{3p_1}}\right)^\frac{3}{2},0, 0 \right) &amp; \text{if } I\sqrt{p_1} \leq \left(\frac{p_2+p_3}{3}\right)^\frac{3}{2}
\end{cases}$$</span></p>

CC BY-SA 4.0
2023-04-28T14:05:13.353Z
2023-11-21T19:55:52.007Z
2023-11-21T19:55:52.007Z
mynameparv
<p>I think that Shiv was in the right direction. Actually, the three points he found satisfy the KKT conditions. If the lagrangean is
<span class="math-container">$$ \mathcal{L} = -x_1^{1/3} - x_2 + \lambda(-I+p_1x_1 + (p_2+p_3)x_2) $$</span>
and the KKT conditions for a minimum are
<span class="math-container">$$ \frac{\partial \mathcal{L}}{\partial x_1} = -\frac13 x_1^{-2/3} + p_1 \lambda \geq 0, x_1\geq 0, \frac{\partial \mathcal{L}}{\partial x_1} x_1 = 0 $$</span>
<span class="math-container">$$ \frac{\partial \mathcal{L}}{\partial x_2} = -1 + (p_2 + p_3) \lambda \geq 0, x_2\geq 0, \frac{\partial \mathcal{L}}{\partial x_2} x_2 = 0 $$</span>
<span class="math-container">$$ \frac{\partial \mathcal{L}}{\partial \lambda} =-I + p_1x_1 + (p_2+p_3)x_2 \leq 0, \lambda\geq 0, \frac{\partial \mathcal{L}}{\partial \lambda} \lambda = 0 $$</span></p>
<p>From the three cases we end up with three points
<span class="math-container">$$ (x_1,x_2) = \left(\left(\frac{3p_1}{p_2+p_3}\right)^{2/3}, I-3^{2/3} \left(\frac{p_1}{p_2+p_3} \right)^{5/3} \right) $$</span>
<span class="math-container">$$ (x_1,x_2) = \left(0, \frac{I}{p_2+p_3} \right) $$</span>
<span class="math-container">$$ (x_1,x_2) = \left(\frac{I}{p_1}, 0 \right) $$</span></p>
<p>But since the agent enjoys <span class="math-container">$x_2$</span> more than <span class="math-container">$x_1$</span> as the latter is cubic rooted, the minimum (or the maximum of the original function) will be choosing the highest <span class="math-container">$x_2$</span>. Hence, the second point is the global minimum. We can assert this is a global minimum because of Weierstrass Theorem. The feasible region is a closed and bounded set, then the utility function (or the negative of) will achieve a maximum (minimum) value over this set.</p>

CC BY-SA 4.0
2023-05-01T02:27:36.133Z
2023-05-01T02:27:36.133Z
Pekisch
<p>Question: An agent who consumes three commodities has a utility function given by:</p>

<p><span class="math-container">$u(x_1,x_2,x_3)=x^{1/3}_1+\min\{ x_2,x_3\}$</span></p>

<p>Given an income <span class="math-container">$I$</span>, and prices of <span class="math-container">$p_1,p_2,p_3$</span>. Describe the consumer’s utility maximization problem. Can the weierstrass and the Kuhn-Tucker theorems be used to obtain and characterize a solution? Why or why not?</p>

<p>attempt: I assume <span class="math-container">$x_i$</span> represents the quantity and belongs to <span class="math-container">$\mathbb R_{+}$</span>. You can form the constraints as follows:
<span class="math-container">$$
x_i \geq0 \quad\forall i \in [3] \\
\sum_{i=1}^3p_ix_i \leq I
$$</span>
You can simplify the objective by noting that for the utility to be the maximum. Hence, the final problem becomes,</p>

<p><span class="math-container">$$\
\max_{x_1, x_2, x_3}x_1^{1/3} +x_2 \quad s.t. \\
x_i \geq0 \quad\forall i \in [3] \\
\sum_{i=1}^3p_ix_i \leq I \\
x_2=x_3
$$</span>
Let's eliminate <span class="math-container">$x_3$</span> as we know that <span class="math-container">$x_2=x_3$</span>. The problem simplifies to
<span class="math-container">$$\
\min_{x_1,\ x_2}\ -x_1^{1/3} -x_2 \quad s.t. \\
x_1 \geq0,\ x_2 \geq 0 \\
p_1x_1 + (p_2 + p_3)x_2 \leq I
$$</span>
<span class="math-container">$$\
\mathcal L(x_1, x_2)=-x_1^{1/3} -x_2 + \lambda_1(-x_1) + \lambda_2(-x_2) + \lambda(p_1x_1 + (p_2 + p_3)x_2 - I)
$$</span></p>

<p>Comment: I am unsure how to take this further. I keep messing up the derivatives ( I assume) and when I try to solve for lambda I manage to fail to isolate the lambda variable let alone getting the variables x1,x2,x3. My professor encouraged me to try this complex problem as an “exercise for the reader.” How do I carry this further, or can someone show me a step by step solution from this point onwards?</p>

CC BY-SA 4.0
2020-06-13T22:10:07.430Z
Hi: I must be missing something but why does $x_2 = x_3$. Is that a given constraint ?
mark leeds
CC BY-SA 4.0
2020-06-13T22:11:11.200Z
Basically $x_2=x_3 $ because that’s the only way a max or min can hold.
Tony456
CC BY-SA 4.0
2020-06-14T00:35:15.980Z
"by noting that for the utility to be the maximum." - something seems to be missing here.
VARulle
CC BY-SA 4.0
2020-06-14T01:46:39.123Z
I'm not questioning you but just trying to understand. Are you saying that the only way to maximize the minimum of $x_2$ and $x_3$ is to set them equal ? If so, I get it. Thanks.
mark leeds
CC BY-SA 4.0
2020-06-14T03:00:48.173Z
Hello, yes, that is what I am saying. We are given nothing else about the two variables and their relationship and as such the minimum of both of them have to the same number.
Tony456
CC BY-SA 4.0
2020-06-14T06:36:00.197Z
This question was [crossposted](https://math.stackexchange.com/questions/3718551/can-the-weierstra%c3%9f-and-or-kuhn-tucker-theorems-be-used-to-obtain-and-characteriz).
Giskard
<mathematical-economics><optimization>
Can the weierstrass and the Kuhn-Tucker theorems be used to obtain and characterize a solution? Why or why not?
Tony456
# Can the weierstrass and the Kuhn-Tucker theorems be used to obtain and characterize a solution? Why or why not?

(Asked by: username_0 on 2020-06-13)

Question: An agent who consumes three commodities has a utility function given by:

$u(x\_1,x\_2,x\_3)=x^{1/3}\_1+\min\{ x\_2,x\_3\}$

Given an income $I$, and prices of $p\_1,p\_2,p\_3$. Describe the consumer’s utility maximization problem. Can the weierstrass and the Kuhn-Tucker theorems be used to obtain and characterize a solution? Why or why not?

attempt: I assume $x\_i$ represents the quantity and belongs to $\mathbb R\_{+}$. You can form the constraints as follows:
$$
x\_i \geq0 \quad\forall i \in [3] \\
\sum\_{i=1}^3p\_ix\_i \leq I
$$
You can simplify the objective by noting that for the utility to be the maximum. Hence, the final problem becomes,

$$\
\max\_{x\_1, x\_2, x\_3}x\_1^{1/3} +x\_2 \quad s.t. \\
x\_i \geq0 \quad\forall i \in [3] \\
\sum\_{i=1}^3p\_ix\_i \leq I \\
x\_2=x\_3
$$
Let's eliminate $x\_3$ as we know that $x\_2=x\_3$. The problem simplifies to
$$\
\min\_{x\_1,\ x\_2}\ -x\_1^{1/3} -x\_2 \quad s.t. \\
x\_1 \geq0,\ x\_2 \geq 0 \\
p\_1x\_1 + (p\_2 + p\_3)x\_2 \leq I
$$
$$\
\mathcal L(x\_1, x\_2)=-x\_1^{1/3} -x\_2 + \lambda\_1(-x\_1) + \lambda\_2(-x\_2) + \lambda(p\_1x\_1 + (p\_2 + p\_3)x\_2 - I)
$$

Comment: I am unsure how to take this further. I keep messing up the derivatives ( I assume) and when I try to solve for lambda I manage to fail to isolate the lambda variable let alone getting the variables x1,x2,x3. My professor encouraged me to try this complex problem as an “exercise for the reader.” How do I carry this further, or can someone show me a step by step solution from this point onwards?

**Tags:** mathematical-economics, optimization
> **Comments:**
> **username_1**: Hi: I must be missing something but why does $x_2 = x_3$. Is that a given constraint ?  
> **username_0**: Basically $x_2=x_3 $ because that’s the only way a max or min can hold.  
> **username_2**: "by noting that for the utility to be the maximum." - something seems to be missing here.  
> **username_1**: I'm not questioning you but just trying to understand. Are you saying that the only way to maximize the minimum of $x_2$ and $x_3$ is to set them equal ? If so, I get it. Thanks.  
> **username_0**: Hello, yes, that is what I am saying. We are given nothing else about the two variables and their relationship and as such the minimum of both of them have to the same number.  
> **username_3**: This question was [crossposted](https://math.stackexchange.com/questions/3718551/can-the-weierstra%c3%9f-and-or-kuhn-tucker-theorems-be-used-to-obtain-and-characteriz).  

---

## 3 answers

### Answer by username_4 on 2020-06-14. Score: 2

I assume $x\_i$ represents the quantity and belongs to $\mathbb R\_{+}$. You can form the constraints as follows:
$$
x\_i \geq0 \quad\forall i \in [3] \\
\sum\_{i=1}^3p\_ix\_i \leq I
$$
You can simplify the objective by noting that for the utility to be the maximum, $x\_2 =x\_3$. Try to reason why this is true. Hence, the final problem becomes,

$$\
\max\_{x\_1, x\_2, x\_3}x\_1^{1/3} +x\_2 \quad s.t. \\
x\_i \geq0 \quad\forall i \in [3] \\
\sum\_{i=1}^3p\_ix\_i \leq I \\
x\_2=x\_3
$$
Can you now set up the lagrange multiplier function? As far as KKT conditions are concerned, I believe they should help because you are trying to maximize a concave function(in $\mathbb R\_+$) subject to convex constraints and the strict inequality constraint is linear.

**Edit 1:** Let's eliminate $x\_3$ as we know that $x\_2=x\_3$. The problem simplifies to
$$\
\min\_{x\_1,\ x\_2}\ -x\_1^{1/3} -x\_2 \quad s.t. \\
x\_1 \geq0,\ x\_2 \geq 0 \\
p\_1x\_1 + (p\_2 + p\_3)x\_2 \leq I
$$
$$\
\mathcal L(x\_1, x\_2)=-x\_1^{1/3} -x\_2 + \lambda\_1(-x\_1) + \lambda\_2(-x\_2) + \lambda(p\_1x\_1 + (p\_2 + p\_3)x\_2 - I)
$$
**Edit 2:** I see solving the Lagrangian can be tough. Let's simplify, we know that all constraints cannot be active at the same time. It will yield an utility of zero.

**Case-1:** $x\_1=0, x\_2>0$
You can check algebraically, $x\_2=\frac{I}{p\_2+p\_3}$. Hence, utility$(U)=\frac{I}{p\_2+p\_3}$

**Case-2:** $x\_1>0, x\_2=0$
You can check algebraically, $x\_1=\frac{I}{p\_1}$. Hence, utility$(U)=\left(\frac{I}{p\_1}\right)^{1/3}$

**Case-3** $x\_1>0, x\_2>0$
You can now use the Lagrangian, its much simpler than the original Lagrangian problem because $\lambda\_1=\lambda\_2=0$
$$\
\mathcal L(x\_1, x\_2)=-x\_1^{1/3} -x\_2 + \lambda(p\_1x\_1 + (p\_2 + p\_3)x\_2 - I)\\
\frac{\partial\mathcal L}{\partial x\_1} = 0 \implies x\_1 = \left(\frac{1}{3\lambda p\_1}\right)^{3/2}\\
\frac{\partial\mathcal L}{\partial x\_2} = 0 \implies \lambda = \frac{1}{p\_2+p\_3}\\
\text{Use the constraint, }p\_1x\_1 + (p\_2 + p\_3)x\_2 = I \text{ to find } x\_2
$$
You can simply finish saying the solution is $\max$ of all three cases. As per the KKT conditions are concerned, one of the three solutions will satisfy KKT Conditions depending on $p\_1, p\_2, p\_3 \text{ and } I$. Note that this is no different from the original problem and hence, KKT condition should be satisfied by one of the three depending on $p\_1, p\_2, p\_3 \text{ and } I$. We could have obtained the same solution using the original Lagrangian. Even there we would have ended up with a $\max$ because of $\lambda\_1, \lambda\_2$ and invoking KKT.


---

### Answer by username_5 on 2023-04-28. Score: 0

I try to use substitution instead of relying on KKT or the Lagrangian method until I encounter a problem that is better solved through them. In a sense, it is like people preferring different methods like elimination or substitution for solving a simple system of equations, but here we are talking about optimization problems instead of systems of equations

We can solve the problem you provided using substitution and simplifying the optimization problem.

Our UMP in consideration is:
$$\begin{aligned}
&\max\_{x\_1,x\_2,x\_3\geq0} \quad  x\_1^\frac{1}{3} + \min(x\_2,x\_3)\\
&\textrm{s.t.} \quad  p\_1x\_1+p\_2x\_2+p\_3x\_3\leq I \end{aligned}$$

Now, what I can do is first solve the UMP holding $x\_1$ constant to get demands for $x\_2, x\_3$ as functions of $x\_1$ and other parameters:
$$\begin{aligned}
& \max\_{x\_1}  {\begin{aligned} \begin{pmatrix} \displaystyle\max\_{x\_2,x\_3} & x\_1^\frac{1}{3}+\min(x\_2,x\_3) \\ \textrm{s.t.}  & p\_2x\_2+p\_3x\_3\leq I-p\_1x\_1 \end{pmatrix} \end{aligned}}\\ \end{aligned}$$
Notice that UMP involving $x\_2,x\_3$ is a standard problem over perfect complements. The problem inside the brackets gives us:
$$x\_2(x\_1,p,I)=x\_3(x\_1,p,I)=\frac{I-p\_1x\_1}{p\_2+p\_3}$$

We can now use the above, substitute them into the problem inside the bracket of the original UMP and solve the optimization to get optimal $x\_1^\*$:
$$\begin{aligned}
&\max\_{0 \leq x\_1\leq\frac{I}{p\_1}} \quad x\_1^\frac{1}{3}+\frac{I-p\_1x\_1}{p\_2+p\_3}\\
\text{gives} \quad & x\_1^\*(p,I)=\left({\frac{p\_2+p\_3}{3p\_1}}\right)^\frac{3}{2}
\end{aligned}$$

Substituting $x\_1^\*$ in the expression for $x\_2,x\_3$ that we derived before gives us the final demands:

$$(x\_1^\*,x\_2^\*,x\_3^\*)(p,I)=\begin{cases}
\left (\left({\frac{p\_2+p\_3}{3p\_1}}\right)^\frac{3}{2},\frac{I}{p\_2+p\_3}-\frac{1}{3}\left(\frac{p\_2+p\_3}{3p\_1}\right)^\frac{1}{2}, \frac{I}{p\_2+p\_3}-\frac{1}{3}\left(\frac{p\_2+p\_3}{3p\_1}\right)^\frac{1}{2}\right) & \text{if } I\sqrt{p\_1} > \left(\frac{p\_2+p\_3}{3}\right)^\frac{3}{2}\\
\left (\left({\frac{p\_2+p\_3}{3p\_1}}\right)^\frac{3}{2},0, 0 \right) & \text{if } I\sqrt{p\_1} \leq \left(\frac{p\_2+p\_3}{3}\right)^\frac{3}{2}
\end{cases}$$


---

### Answer by username_6 on 2023-05-01. Score: 0

I think that Shiv was in the right direction. Actually, the three points he found satisfy the KKT conditions. If the lagrangean is
$$ \mathcal{L} = -x\_1^{1/3} - x\_2 + \lambda(-I+p\_1x\_1 + (p\_2+p\_3)x\_2) $$
and the KKT conditions for a minimum are
$$ \frac{\partial \mathcal{L}}{\partial x\_1} = -\frac13 x\_1^{-2/3} + p\_1 \lambda \geq 0, x\_1\geq 0, \frac{\partial \mathcal{L}}{\partial x\_1} x\_1 = 0 $$
$$ \frac{\partial \mathcal{L}}{\partial x\_2} = -1 + (p\_2 + p\_3) \lambda \geq 0, x\_2\geq 0, \frac{\partial \mathcal{L}}{\partial x\_2} x\_2 = 0 $$
$$ \frac{\partial \mathcal{L}}{\partial \lambda} =-I + p\_1x\_1 + (p\_2+p\_3)x\_2 \leq 0, \lambda\geq 0, \frac{\partial \mathcal{L}}{\partial \lambda} \lambda = 0 $$

From the three cases we end up with three points
$$ (x\_1,x\_2) = \left(\left(\frac{3p\_1}{p\_2+p\_3}\right)^{2/3}, I-3^{2/3} \left(\frac{p\_1}{p\_2+p\_3} \right)^{5/3} \right) $$
$$ (x\_1,x\_2) = \left(0, \frac{I}{p\_2+p\_3} \right) $$
$$ (x\_1,x\_2) = \left(\frac{I}{p\_1}, 0 \right) $$

But since the agent enjoys $x\_2$ more than $x\_1$ as the latter is cubic rooted, the minimum (or the maximum of the original function) will be choosing the highest $x\_2$. Hence, the second point is the global minimum. We can assert this is a global minimum because of Weierstrass Theorem. The feasible region is a closed and bounded set, then the utility function (or the negative of) will achieve a maximum (minimum) value over this set.


---

