Which is the linear system of equations?

[mathmari]

Hey!

A national economy has $3$ sectors: fishing, forestry, boat building.

One fishing boat is needed, to catch two tonnes of fish.
To produce four tonnes of wood, one tonne of fish is needed to feed the forestry workers.
Two tonnes of wood are needed, to build one fishing boat.

There is also an external demand for $n_1$ tonnes of fish, for $n_2$ tonnes of wood and for $n_3$ fishing boats.

$x_1$ is the total tonnes of fish that have to be producted, $x_2$ is the total tonnes of wood and $x_3$ is the total number of fishing boats that have to be produced. How can we tranform this into a linear system of equations? Could you give me a hint? (Wondering)

 

[I like Serena]

Hey!

A national economy has $3$ sectors: fishing, forestry, boat building.

One fishing boat is needed, to catch two tonnes of fish.
To produce four tonnes of wood, one tonne of fish is needed to feed the forestry workers.
Two tonnes of wood are needed, to build one fishing boat.

There is also an external demand for $n_1$ tonnes of fish, for $n_2$ tonnes of wood and for $n_3$ fishing boats.

$x_1$ is the total tonnes of fish that have to be producted, $x_2$ is the total tonnes of wood and $x_3$ is the total number of fishing boats that have to be produced. How can we tranform this into a linear system of equations? Could you give me a hint? (Wondering)

Hey mathmari! (Smile)

Suppose we fish $x_1$ tonnes of fish, we produce $x_2$ tonnes of wood, and we build $x_3$ fishing boats.
Which relations will we have between $x_1, x_2$ and $x_3$? (Wondering)

 

[mathmari]

Suppose we fish $x_1$ tonnes of fish, we produce $x_2$ tonnes of wood, and we build $x_3$ fishing boats.
Which relations will we have between $x_1, x_2$ and $x_3$? (Wondering)

For $2$ tonnes of fish, we need $1$ fishing boat. So, to fish $x_1$ tonnes of fish we need $\frac{x_1}{2}$ boats.
For $1$ boat we need $2$ tonnes of wood. Since we need $\frac{x_1}{2}$ boats, we need $x_1$ tonnes of wood.
For $4$ tonnes of wood, we need $1$ tonne of fish. Since we need $x_1$ tonnes of wood, we need $\frac{x_1}{4}$ tonnes of fish.

Is this correct? (Wondering)

 

[I like Serena]

For $2$ tonnes of fish, we need $1$ fishing boat. So, to fish $x_1$ tonnes of fish we need $\frac{x_1}{2}$ boats.

Yep.
So $x_1 \text{ tonnes of fish} \le 2 x_3 \text{ fishing boats}$.

For $1$ boat we need $2$ tonnes of wood. Since we need $\frac{x_1}{2}$ boats, we need $x_1$ tonnes of wood.

Hold on. Wasn't $x_1$ the amount of fish? What does that have to do with boats or wood? (Wondering)

For $4$ tonnes of wood, we need $1$ tonne of fish. Since we need $x_1$ tonnes of wood, we need $\frac{x_1}{4}$ tonnes of fish.

Yes.
So $4x_2 \text{ tonnes of wood} \le x_1 \text{ tonnes of fish}$, yes? (Thinking)

 

[mathmari]

For $2$ tonnes of fish, we need $1$ fishing boat. So, to fish $x_1$ tonnes of fish we need $\frac{x_1}{2}$ boats. So that there exist so many boats it must hold $\frac{x_1}{2}\leq x_3 \Rightarrow x_1\leq 2x_3$. For $1$ boat we need $2$ tonnes of wood. To build $x_3$ boats, we need $2x_3$ tonnes of wood. So that there exist so many tonnes of wood it must hold $2x_3\leq x_2$. For $4$ tonnes of wood, we need $1$ tonne of fish. To produce $x_2$ tonnes of wood, we need $\frac{x_2}{4}$ tonnes of fish. So that there exist so many tonnes of fish it must hold $\frac{x_2}{4}\leq x_1 \Rightarrow x_2\leq 4x_1$.

So $4x_2 \text{ tonnes of wood} \le x_1 \text{ tonnes of fish}$, yes? (Thinking)

How do we get this inequality? What did I wrong above? (Wondering) Are the other inequalities now correct? (Wondering) So, we have inequalities now. To get equalities do we use the external demands? (Wondering)

 

[I like Serena]

How do we get this inequality? What did I wrong above? (Wondering)

You're right. (Blush)

Are the other inequalities now correct? (Wondering)

So, we have inequalities now. To get equalities do we use the external demands? (Wondering)

We should indeed add the external demands.
As I see it that gives a linear system of inequalities that we can optimize based on some objective function.
Since the problem statement asks for equations, perhaps we're supposed to change the inequality signs to equal signs. (Thinking)

 

[mathmari]

For $2$ tonnes of fish, we need $1$ fishing boat. So, to fish $x_1$ tonnes of fish we need $\frac{x_1}{2}$ boats. So that there exist so many boats it must hold $\frac{x_1}{2}\leq x_3 \Rightarrow x_1\leq 2x_3$.

We have that $\frac{x_1}{2}\leq x_3$. That means that of the total number of boats, we need $\frac{x_1}{2}$ to fish the need tonnes of fish and $n_3$ for the external demand.

Do we get then the equality $\frac{x_1}{2}+n_3=x_3$ ?

For $1$ boat we need $2$ tonnes of wood. To build $x_3$ boats, we need $2x_3$ tonnes of wood. So that there exist so many tonnes of wood it must hold $2x_3\leq x_2$.

We have that $2x_3\leq x_2$. That means that of the total tonnes of wood, we need $2x_3$ tonnes to build the needed boats and $n_2$ for the external demand.

Do we get then the equality $2x_3+n_2=x_2$ ?

For $4$ tonnes of wood, we need $1$ tonne of fish. To produce $x_2$ tonnes of wood, we need $\frac{x_2}{4}$ tonnes of fish. So that there exist so many tonnes of fish it must hold $\frac{x_2}{4}\leq x_1 \Rightarrow x_2\leq 4x_1$.

We have that $\frac{x_2}{4}\leq x_1$. That means that of the total tonnes of fish, we need $\frac{x_2}{4}$ tonnes to produce the needed tonnes of wood and $n_1$ for the external demand.

Do we get then the equality $\frac{x_2}{4}+n_1=x_1$ ?(Wondering)

As I see it that gives a linear system of inequalities that we can optimize based on some objective function.
Since the problem statement asks for equations, perhaps we're supposed to change the inequality signs to equal signs. (Thinking)

What do you mean? (Wondering)

 

[I like Serena]

We have that $\frac{x_1}{2}\leq x_3$. That means that of the total number of boats, we need $\frac{x_1}{2}$ to fish the need tonnes of fish and $n_3$ for the external demand.

Do we get then the equality $\frac{x_1}{2}+n_3=x_3$ ?

We have that $2x_3\leq x_2$. That means that of the total tonnes of wood, we need $2x_3$ tonnes to build the needed boats and $n_2$ for the external demand.

Do we get then the equality $2x_3+n_2=x_2$ ?

We have that $\frac{x_2}{4}\leq x_1$. That means that of the total tonnes of fish, we need $\frac{x_2}{4}$ tonnes to produce the needed tonnes of wood and $n_1$ for the external demand.

Do we get then the equality $\frac{x_2}{4}+n_1=x_1$ ?(Wondering)

What do you mean? (Wondering)

Let's run through this from begin to end... (Thinking)

Suppose we have a sufficient total tonnes of fish $x_1$ to which we'll get back later to see if we can actually fish it.
Then we have to provide $n_1$ tonnes of fish to external demand, leaving us with $x_1 - n_1$ tonnes of fish.
We use it to feed to forestry workers who can then build a total of:
$$x_2 = 4(x_1 - n_1) \text{ tonnes of wood} \tag 1$$
We provide $n_2$ to external demand, leaving us with $x_2 - n_2$ tonnes of wood to build a total of:
$$x_3 = \frac 12(x_2 - n_2) \text{ boats} \tag 2$$
With and external demand of $n_3$, that leaves us with $x_3 - n_3$ boats to fish a total of:
$$x_1 = 2(x_3 - n_3) \text{ tonnes of fish} \tag 3$$

If we can make sure that is exactly enough for what we needed to begin with, we have a solution with no waste.
So our system of linear equations is:
$$\begin{cases}x_2 = 4(x_1 - n_1) \\ x_3 = \frac 12(x_2 - n_2) \\ x_1 = 2(x_3 - n_3)\end{cases} \Rightarrow
\begin{cases}4x_1 &- x_2 &&=& 4n_1 \\ & \phantom +x_2 &- 2x_3 &=& \phantom 1n_2 \\ -x_1 &&+ 2x_3 &=& 2n_3\end{cases} \tag 4
$$(Thinking)

 

[mathmari]

Shouldn't the system be \begin{align*}4x_1-x_2 \ \ \ \ \ \ \ \ \ & = 4n_1 \\ \ \ \ \ \ \ \ \ \ x_2-2x_3& = \ n_2 \\ -x_1 \ \ \ \ \ \ \ \ +2x_3& = 2n_3\end{align*} ? (Wondering) If the total outputs of fish, wood und fishin boats are $x_1=160$ tonnes, $x_2=240$ tonnes and $x_3=80$ boats, then I want to find the external demands.

Is the above system is correct, we get the following:
\begin{align*}&4n_1=4x_1-x_2=4\cdot 160-240=640-240=400 \Rightarrow n_1=100 \\ &n_2=x_2-2x_3 =240-2\cdot 80=240-160 \Rightarrow n_2=80 \\ &2n_3=-x_1+2x_3=-160+2\cdot 80=-160+160=0 \Rightarrow n_3=0\end{align*}

Is this correct? (Wondering)

 

[I like Serena]

Shouldn't the system be \begin{align*}4x_1-x_2 \ \ \ \ \ \ \ \ \ & = 4n_1 \\ \ \ \ \ \ \ \ \ \ x_2-2x_3& = \ n_2 \\ -x_1 \ \ \ \ \ \ \ \ +2x_3& = 2n_3\end{align*} ? (Wondering)

Ah yes. I fixed it.

If the total outputs of fish, wood und fishin boats are $x_1=160$ tonnes, $x_2=240$ tonnes and $x_3=80$ boats, then I want to find the external demands.

Is the above system is correct, we get the following:
\begin{align*}&4n_1=4x_1-x_2=4\cdot 160-240=640-240=400 \Rightarrow n_1=100 \\ &n_2=x_2-2x_3 =240-2\cdot 80=240-160 \Rightarrow n_2=80 \\ &2n_3=-x_1+2x_3=-160+2\cdot 80=-160+160=0 \Rightarrow n_3=0\end{align*}

Is this correct? (Wondering)

Let's see...

We have $160 - 100 = 60$ tonnes of fish to feed the forestry workers.
They produce $4\cdot 60 = 240$ tonnes of wood.
Then we have $240 - 80 = 160$ tonnes of wood to build boats.
We can build $\frac 12 \cdot 160 = 80$ boats.
It leaves us with $80-0 = 80$ boats to fish.
So we can fish $2\cdot 80=160$ tonnes, which is exactly what we started with.

So yes, I think it's correct. (Happy)

 

[mathmari]

Great! Thank you very much! (Sun)