How Do You Calculate Ore Production and Costs for Zinc and Aluminium?

[mathmari]

Hey!

A mining company produce in three mines aluminium and zinc. At the following matrix give the information about the amount of zinc and alluminium at a produced tonne of ore and the corresponding cost per tonne of ore:

$$\begin{matrix}
\text{Mine } & \text{ Zinc } & \text{ Alluminium } & \text{ Cost/ tonne of ore }\\
1 & 10\% & 40\% & 60\\
2 & 20\% & 30\% & 45\\
3 & 30\% & 20\% & 40
\end{matrix}$$ For a specific costumer 100 t of zinc and 200 t of alluminium have to be produced. We have to give a system of equations, from which the required amounts for the production at each mine can be computed. We have to give also all the solutions of this system.
I have done the following but I am not really sure if I understood correctly the exercise... (Thinking)

Let $z_i$ be the amount of zinc produced by the mine $i$ and let $a_i$ be the amount of alluminium produced by the mine $i$.

Then do we get the following system?

$$0.1x_1+0.2x_2+0.3x_3=100 \\ 0.4a_1+0.3a_2+0.2a_3=200$$

Is this correct? (Wondering)

 

[I like Serena]

Hey!

A mining company produce in three mines aluminium and zinc. At the following matrix give the information about the amount of zinc and alluminium at a produced tonne of ore and the corresponding cost per tonne of ore:

$$\begin{matrix}
\text{Mine } & \text{ Zinc } & \text{ Alluminium } & \text{ Cost/ tonne of ore }\\
1 & 10\% & 40\% & 60\\
2 & 20\% & 30\% & 45\\
3 & 30\% & 20\% & 40
\end{matrix}$$ For a specific costumer 100 t of zinc and 200 t of alluminium have to be produced. We have to give a system of equations, from which the required amounts for the production at each mine can be computed. We have to give also all the solutions of this system.
I have done the following but I am not really sure if I understood correctly the exercise... (Thinking)

Let $z_i$ be the amount of zinc produced by the mine $i$ and let $a_i$ be the amount of alluminium produced by the mine $i$.

Then do we get the following system?

$$0.1x_1+0.2x_2+0.3x_3=100 \\ 0.4a_1+0.3a_2+0.2a_3=200$$

Is this correct? (Wondering)

Hey mathmari! (Smile)

I think a mine produces tonnes of ore and not specifically zinc or aluminium, but it gets refined to generate percentages of zinc and aluminium.
So I think we should define $x_i$ to be the amount of ore produced in each mine.
That gives us:
$$0.1x_1+0.2x_2+0.3x_3 = 100\text{ t of Zinc} \\ 0.4x_1+0.3x_2+0.2x_3 = 200\text{ t of Aluminium}$$
for which we might find all solutions.

To be fair, we have to produce at least a sufficient amount of each metal - from a practical point of view, there's nothing wrong with it being more.
So we should actually have the inequalities:
$$0.1x_1+0.2x_2+0.3x_3 \ge 100 \text{ t of Zinc}\\ 0.4x_1+0.3x_2+0.2x_3\ge 200 \text{ t of Aluminium}$$
which we might solve as well.

And it's not mentioned (yet), but presumably we will want to minimize the cost.
So we'd get the expression:
$$\text{Cost } = 60 x_1 + 45 x_2 + 40 x_3$$
that we want to minimize, giving us an optimal solution instead of all possible solutions.

 

[mathmari]

I think a mine produces tonnes of ore and not specifically zinc or aluminium, but it gets refined to generate percentages of zinc and aluminium.
So I think we should define $x_i$ to be the amount of ore produced in each mine.
That gives us:
$$0.1x_1+0.2x_2+0.3x_3 = 100\text{ t of Zinc} \\ 0.4x_1+0.3x_2+0.2x_3 = 200\text{ t of Aluminium}$$
for which we might find all solutions.

Ah ok... (Thinking)

So, do we have the following: For each mine $i$ we have that each tonne of ore is $z_i\%$ Zinc and $y_i\%$ aluminium, for $z_i\in \{10, 20, 30\}$ and $y_i\in \{40, 30, 20\}$.

So, when we produce in the mine $i$ $\ \ x_i$ tonnes of ore we have $z_ix_i$ zinc and $y_ix_i$ aluminium.

We suppose that each mine produce $x_i$ tonnes of ore and we want to compute the amount of zinc the that of aluminium.

So, we have the following:

\begin{equation*}\begin{matrix}
\text{Mine } & \text{ Zinc } x_i & \text{ Aluminium } y_i & \text{ Costs/tonne of ore }\\
1 & 10\% & 40\% & 60\\
2 & 20\% & 30\% & 45\\
3 & 30\% & 20\% & 40
\end{matrix}\end{equation*}

We suppose that the mines $1,2$ and $3$ produce $ x_1, x_2$ and $x_3$ tonnes of ore.

Therefore, we can write the table also as follows:

\begin{equation*}\begin{matrix}
\text{Tonnes produced in mine}\ i & \text{ Amount of zinc } & \text{ Amount of Aluminium } & \text{ Total costs of the production}\\
x_1 & 0.1x_1 & 0.4x_1 & x_160\\
x_2 & 0.2x_2 & 0.3x_2 & x_245\\
x_3 & 0.3x_3 & 0.2x_3 & x_340
\end{matrix}\end{equation*}

We want that $100$ tonnes of zinc and $200$ tonnes of aluminium are produced.

So, we get the following system of equations:

$$0.1x_1 + 0.2x_2 + 0.3x_3 = 100\\ 0.4x_1 + 0.3x_2 + 0.2x_3 = 200$$

We solve this system using the Gauss Algorithm:
$$\begin{bmatrix}
\begin{matrix}
0.1 & 0.2 & 0.3\\
0.4 & 0.3 & 0.2
\end{matrix}\left|\begin{matrix}
100\\
200
\end{matrix}\right.\end{bmatrix}\begin{matrix}
\\
2.\text{row}- 4\cdot 1.\text{row}
\end{matrix} \longrightarrow \begin{bmatrix}
\begin{matrix}
0.1 & 0.2 & 0.3\\
0 & -0.5 & 1
\end{matrix}\left|\begin{matrix}
100\\
-200
\end{matrix}\right.\end{bmatrix}$$ So, we get the following system of equations:
$$0.1x_1+0.2x_2+0.3x_3=100 \\ -0.5x_2+x_3=-200$$
From the second we get $x_3=0.5t_2-200$.
From the fist we get then $0.1x_1+0.2x_2+0.3 (0.5x_2-200)=100 \Rightarrow 0.1x_1=100-0.2x_2-1.5x_2+60\Rightarrow 0.1x_1=-0.35x_2+160 \Rightarrow x_1=-3.5x_2+1600$.

So, all the solutions of the system of equations are $$ (x_1, x_2, x_3)=(3.5x_2+1600, x_2, 0.5x_2-200)=x_2(3.5, 1, 0.5)+(1600, 0, -200), \ x_2\in \mathbb{R}.$$

Is everything correct so far? (Wondering)

To be fair, we have to produce at least a sufficient amount of each metal - from a practical point of view, there's nothing wrong with it being more.
So we should actually have the inequalities:
$$0.1x_1+0.2x_2+0.3x_3 \ge 100 \text{ t of Zinc}\\ 0.4x_1+0.3x_2+0.2x_3\ge 200 \text{ t of Aluminium}$$
which we might solve as well.

Is it also correct to solve the equalities instead of the inequalities? (Wondering)

And it's not mentioned (yet), but presumably we will want to minimize the cost.
So we'd get the expression:
$$\text{Cost } = 60 x_1 + 45 x_2 + 40 x_3$$
that we want to minimize, giving us an optimal solution instead of all possible solutions.

The second question is the following:

At the table we can see also the costs for the production and the separation of the metals. What amounts does it have to be produced in each mine so that the costs are minimal? Which are the minimal costs? Which is the best method to use to minimize the Cost? (Wondering)

 

[I like Serena]

Is everything correct so far? (Wondering)

It looks about right to me.

Is it also correct to solve the equalities instead of the inequalities? (Wondering)

To practice solving equations, sure, to really solve the problem, it doesn't really make sense. (Nerd)
Btw, there is another practical condition: all $x_i \ge 0$, because we can't generate a negative amount of ore in a mine.
It seems to me that at least that should be added.

The second question is the following:

At the table we can see also the costs for the production and the separation of the metals. What amounts does it have to be produced in each mine so that the costs are minimal? Which are the minimal costs?

Which is the best method to use to minimize the Cost? (Wondering)

The Simplex algorithm.
The whole problem is a classical example where that algorithm should be used.
It's probably on the next page. (Nerd)

 

[mathmari]

It looks about right to me.

(Nerd)

To practice solving equations, sure, to really solve the problem, it doesn't really make sense. (Nerd)
Btw, there is another practical condition: all $x_i \ge 0$, because we can't generate a negative amount of ore in a mine.
It seems to me that at least that should be added.

Ah ok... (Thinking)

The Simplex algorithm.
The whole problem is a classical example where that algorithm should be used.
It's probably on the next page. (Nerd)

Ah ok! (Nerd) (Thinking)

Considering the system of equations we got that all the solutions are $$(x_1, x_2, x_3)=(3.5x_2+1600, x_2, 0.5x_2-200), \ x_2\geq 0, 0.5x_2\geq 200 \\ \Rightarrow (x_1, x_2, x_3)=(3.5x_2+1600, x_2, 0.5x_2-200), \ x_2\geq 0, x_2\geq 400$$

Could we maybe use also these solutions to minimize the Cost? (Wondering) We have the following: $$\text{Cost } = 60 x_1 + 45 x_2 + 40 x_3=60(3.5x_2+1600)+45x_2+40(0.5x_2-200)=275x_2+88000, x_2\geq 400$$

Since $x_2\geq 400$ and every other term is positive, the cost is minimal when $x_2=400$.
So, the cost is minimal when $(x_1, x_2, x_3)=(1400+1600, 400, 0)=(3000,400,0)$, and the minimal cost is $\text{Cost } = 110000+88000=198000$.

Is this correct? (Wondering)

 

[I like Serena]

Considering the system of equations we got that all the solutions are $$(x_1, x_2, x_3)=(3.5x_2+1600, x_2, 0.5x_2-200), \ x_2\geq 0, 0.5x_2\geq 200 \\ \Rightarrow (x_1, x_2, x_3)=(3.5x_2+1600, x_2, 0.5x_2-200), \ x_2\geq 0, x_2\geq 400$$

Could we maybe use also these solutions to minimize the Cost? (Wondering) We have the following: $$\text{Cost } = 60 x_1 + 45 x_2 + 40 x_3=60(3.5x_2+1600)+45x_2+40(0.5x_2-200)=275x_2+88000, x_2\geq 400$$

Since $x_2\geq 400$ and every other term is positive, the cost is minimal when $x_2=400$.
So, the cost is minimal when $(x_1, x_2, x_3)=(1400+1600, 400, 0)=(3000,400,0)$, and the minimal cost is $\text{Cost } = 110000+88000=198000$.

Is this correct? (Wondering)

Yep. It's correct.
Do consider that the solution $(x_1, x_2, x_3)=(0,666 \frac 23,0)$ satisfies the inequalities, lowers the cost to $30000$, and gives us more Zinc to sell.
Not bad eh? (Wondering)

 

[mathmari]

Do consider that the solution $(x_1, x_2, x_3)=(0,666 \frac 23,0)$ satisfies the inequalities, lowers the cost to $30000$, and gives us more Zinc to sell.
Not bad eh? (Wondering)

(Thinking) (Wondering)

How did you find this? By applying the Simplex Algorithm? (Wondering) At my solution we have the following:
In the mine $1$ $\ 3000\cdot 0.1=300$ tonnes of zirc and $3000\cdot 0.4=1200$ tonnes of aluminium will be produced, in the mine $2$ $\ 400\cdot 0.2=80$ tonnes of zirc and $400\cdot 0.3=120$ tonnes of aluminium and in the mine $3$ $\ 0\cdot 0.3=0$ tonnes of zirc and $0\cdot 2=0$ tonnes of aluminium.

Right? (Wondering)

How does it come that although I took the equations the total number of tonnes of zirc is more than $100$ and the total number of tonnes of aluminium is more than $200$ ? (Wondering)

Have I done something wrong? (Wondering)

 

[I like Serena]

How did you find this? By applying the Simplex Algorithm? (Wondering)

Yep. Excel has it built-in. (Nod)

At my solution we have the following:
In the mine $1$ $\ 3000\cdot 0.1=300$ tonnes of zirc and $3000\cdot 0.4=1200$ tonnes of aluminium will be produced, in the mine $2$ $\ 400\cdot 0.2=80$ tonnes of zirc and $400\cdot 0.3=120$ tonnes of aluminium and in the mine $3$ $\ 0\cdot 0.3=0$ tonnes of zirc and $0\cdot 2=0$ tonnes of aluminium.

Right? (Wondering)

How does it come that although I took the equations the total number of tonnes of zirc is more than $100$ and the total number of tonnes of aluminium is more than $200$ ? (Wondering)

Have I done something wrong? (Wondering)

Ah. It is not a proper solution. Previously I didn't check properly.
Apparently when we check for equality there is no solution at all.

 

[mathmari]

Ah. It is not a proper solution. Previously I didn't check properly.
Apparently when we check for equality there is no solution at all.

So, is it not possible that $100$ t zinc and $200$ t aluminium are produced? (Wondering)
Why is there no solution when we check for equality? (Wondering)

 

[I like Serena]

So, is it not possible that $100$ t zinc and $200$ t aluminium are produced? (Wondering)
Why is there no solution when we check for equality? (Wondering)

We solve this system using the Gauss Algorithm:
$$\begin{bmatrix}
\begin{matrix}
0.1 & 0.2 & 0.3\\
0.4 & 0.3 & 0.2
\end{matrix}\left|\begin{matrix}
100\\
200
\end{matrix}\right.\end{bmatrix}\begin{matrix}
\\
2.\text{row}- 4\cdot 1.\text{row}
\end{matrix} \longrightarrow \begin{bmatrix}
\begin{matrix}
0.1 & 0.2 & 0.3\\
0 & -0.5 & 1
\end{matrix}\left|\begin{matrix}
100\\
-200
\end{matrix}\right.\end{bmatrix}$$

Shouldn't that be:
\begin{bmatrix}
\begin{matrix}
0.1 & 0.2 & 0.3\\
0 & -0.5 & {\color{red}-1}
\end{matrix}\left|\begin{matrix}
100\\
-200
\end{matrix}\right.\end{bmatrix}
(Wondering)

 

[mathmari]

Shouldn't that be:
\begin{bmatrix}
\begin{matrix}
0.1 & 0.2 & 0.3\\
0 & -0.5 & {\color{red}-1}
\end{matrix}\left|\begin{matrix}
100\\
-200
\end{matrix}\right.\end{bmatrix}
(Wondering)

Oh yes... (Tmi)

So, do we have the following?

$$0.1x_1+0.2x_2+0.3x_3=100 \\ -0.5x_2-x_3=-200$$
Grom the second equation we get $x_3=-0.5x_2+200$.
From the first we get then $0.1x_1+0.2x_2+0.3 (-0.5x_2+200)=100 \Rightarrow 0.1x_1=100-0.2x_2+0.15x_2-60\Rightarrow 0.1x_1=-0.05x_2+40 \Rightarrow x_1=-0.5x_2+400$.

Therefore, the solutions are $$(x_1, x_2, x_3)=(-0.5x_2+400, x_2, -0.5x_2+200)$$
where $-0.5x_2+400\geq 0\Rightarrow 0.5x_2\leq 400 \Rightarrow x_2\leq 800$, $x_2\geq 0$ and $-0.5x_2+200\geq 0 \Rightarrow 0.5x_2\leq 200 \Rightarrow x_2\leq 400$.
So, it must be $0\leq x_2\leq 400$.

The cost is $$\text{ Cost } =60x_1+45x_2+40x_3=60(-0.5x_2+400)+45x_2+40(-0.5x_2+200)=32000-5x_2, \ 0\leq x_2\leq 400 $$

$32000-5x_2$ has its minimal value when $x_2$ gets its maximal value. So, we have that the cost is minimal when $x_2=400$,so when $(x_1, x_2, x_3)=(200, 400, 0). $
That means that in the mine $1$ $\ 200\cdot 0.1=20$ tonnes of zirc and $200\cdot 0.4=80$ tonnes of aluminium will be produced, in the mine $2$ $\ 400\cdot 0.2=80$ tonnes of zirc and $400\cdot 0.3=120$ tonnes of aluminium and in the mine $3$ $\ 0\cdot 0.3=0$ tonnes of zirc and $0\cdot 2=0$ tonnes of aluminium.
The minimal cost is $$\text{ Cost } =32000-5\cdot 400=3000 $$

Is it correct now? (Wondering)

 

[I like Serena]

Yep. (Nod)
That is, if we make it $\text{ Cost } =32000-5\cdot 400={\color{red}30000}$.

 

[mathmari]

Yep. (Nod)
That is, if we make it $\text{ Cost } =32000-5\cdot 400={\color{red}30000}$.

Oh yes... (Blush) (Tmi)

Thank you very much! (Happy)