MHB What is the maximum of the given function in the closed square $D$?

[evinda]

Hello! (Wave)

We consider the function $f(x_1,x_2)=x_1+x_2$ and the (closed) square $D$ with edges $(0,0), (1,0), (1,1), (0,1)$.

Check if $f$ has a maximum in $D$ and if so, find the points of $D$ at which this maximum is achieved.$$\max_{(x_1,x_2) \in D} (x_1+x_2)$$

View attachment 4792

$$(S)\left\{\begin{matrix}
x_1 \geq 0\\
x_2 \geq 0\\
x_1 \leq 1\\
x_2 \leq 1
\end{matrix}\right.$$

so $D=\{ (x_1, x_2) \in \mathbb{R}^2: x_1, x_2 \text{ satisfy } (S)\}$.

$$x_1, x_2 \geq 0$$

Obviously $\max_{(x_1, x_2) \in D} (x_1+x_2) \leq \max_{(x_1, x_2) \in D} x_1 + \max_{(x_1, x_2) \in D} x_2 \leq 1+1=2 $

and $\max_{(x_1, x_2) \in D} (x_1+x_2)=f(1,1)$, so $(1,1)$ is the optimal solution of the linear programming problem.
How do we deduce that the maximum of the given function is $2$. Because of the fact that the maximum is less or equal to $2$ and there is a point $(1,1)$ for which the value $2$ is achieved?

 

[I like Serena]

Hello! (Wave)

and $\max_{(x_1, x_2) \in D} (x_1+x_2)=f(1,1)$, so $(1,1)$ is the optimal solution of the linear programming problem.

Small correction: it's "an" optimal solution.
We don't know yet if it is "the" optimal solution. (Nerd)

How do we deduce that the maximum of the given function is $2$. Because of the fact that the maximum is less or equal to $2$ and there is a point $(1,1)$ for which the value $2$ is achieved?

Yes. (Smile)

Still, I believe the question asks for "all" points... (Sweating)

 

[evinda]

Small correction: it's "an" optimal solution.
We don't know yet if it is "the" optimal solution. (Nerd)

Why? Aren't all the other values that $f$ could get smaller? Or am I wrong? (Thinking)

Yes. (Smile)

Still, I believe the question asks for "all" points... (Sweating)

So in order to show that we have one optimal solution do we have to reject the other feasible solutions?

 

[I like Serena]

Why? Aren't all the other values that $f$ could get smaller? Or am I wrong? (Thinking)

So in order to show that we have one optimal solution do we have to reject the other feasible solutions?

Indeed. And that's exactly the point.
Because all other values that $f$ could get are smaller, it is "the" maximum. (Happy)

Moveover, if the boundary were not included, there would be no maximum, only a supremum. (Nerd)

 

[evinda]

Indeed. And that's exactly the point.
Because all other values that $f$ could get are smaller, it is "the" maximum. (Happy)

Moveover, if the boundary were not included, there would be no maximum, only a supremum. (Nerd)

In order to reject the other feasible solutions do we have to compute the values $f(0,0), f(1,0), f(0,1)$ and the value of $f\left( \frac{1}{m}, \frac{1}{n}\right), n,m \in \mathbb{N}$ ? (Thinking)

 

[I like Serena]

In order to reject the other feasible solutions do we have to compute the values $f(0,0), f(1,0), f(0,1)$ and the value of $f\left( \frac{1}{m}, \frac{1}{n}\right), n,m \in \mathbb{N}$ ? (Thinking)

From a linear programming point of view, indeed we need to check all the corner points.
But that only works if the function to maximize is a linear function.

Where do your $f\left( \frac{1}{m}, \frac{1}{n}\right)$ come from? (Wondering)

More generally, we would take the partial derivatives and see where they are zero.
Additionally, we would inspect the boundaries to see if there is a maximum there.
For linear programming problems that can be simplified to just looking at the corner points. (Nerd)

 

[evinda]

From a linear programming point of view, indeed we need to check all the corner points.
But that only works if the function to maximize is a linear function.

Where do your $f\left( \frac{1}{m}, \frac{1}{n}\right)$ come from? (Wondering)

More generally, we would take the partial derivatives and see where they are zero.
Additionally, we would inspect the boundaries to see if there is a maximum there.
For linear programming problems that can be simplified to just looking at the corner points. (Nerd)

Ah I didn't know that if we have a linear programming problem we look only at the corner points. (Whew)

So we say that $\max_{(x_1, x_2) \in D} (x_1+x_2) \leq 2$ and

$$f(0,0)=0 \\ f(1,0)=1 \\ f(1,1)=2 \\ f(0,1)=1$$

thus the maximum of the function $f$ is $2$ and is achieved for $(x_1, x_2)=(1,1)$, right?

 

[I like Serena]

Ah I didn't know that if we have a linear programming problem we look only at the corner points. (Whew)

So we say that $\max_{(x_1, x_2) \in D} (x_1+x_2) \leq 2$ and

$$f(0,0)=0 \\ f(1,0)=1 \\ f(1,1)=2 \\ f(0,1)=1$$

thus the maximum of the function $f$ is $2$ and is achieved for $(x_1, x_2)=(1,1)$, right?

Yep! (Happy)

 

[evinda]

Yep! (Happy)

Nice... Thank you! (Smile)