#### Hendersonc241

##### New member

- Joined
- Mar 18, 2021

- Messages
- 9

- Thread starter Hendersonc241
- Start date

- Joined
- Mar 18, 2021

- Messages
- 9

- Joined
- Nov 12, 2017

- Messages
- 11,537

This isn't clear to me. First you say you want to "determine the

How about giving us a simple

- Joined
- Mar 18, 2021

- Messages
- 9

\(\displaystyle v(t) = x(t) + p_1(t)m_1 + p_2(t)m_2 + p_3(t)m_3.\)

What that means is that the value of the item at time t is some -non-negative value x greater than the sum of the values of the three constituent metals.

If you assume that x is constant over time, you can figure out what the quantities of metals are if you have numerical data for v, p_1, p_2, and p_3 on four different dates.

But that may not be a very good assumption.

- Joined
- Mar 18, 2021

- Messages
- 9

- Joined
- Nov 12, 2017

- Messages
- 11,537

Is this a real-life problem, or an assignment?

If it's the latter, and you don't have to think about issues like whether the value depends on other things besides the values of the three metals (such as manufacturing costs or other materials), then you can just write and solve a system of three linear equations using the three sets of numbers you gave.

If it's real-life, then you'll have to make further assumptions, such as JeffM's, which adds a fourth variable. You can use additional values beyond four to test for consistency of your assumptions.

I'm not sure of your objection. Are you assuming that the value of the item is affected only by the three metals, and nothing else? The suggested "x" is not the amount of metals, but the "overhead", in some sense, which I'd think would have to exist. But then, I know nothing about the real "item".

Of course the weight of the unit exceeds the weight of some of its materials. But that is not what the equation is about.

You seem to be assuming that the price at which an item is exchanged equals the cost of the materials. This is generally not even close to correct. Do you think that the price of a car is simply the sum of the prices of the parts. In fact, the greatest component of the price of any physical good is almost always labor cost, not cost of materials.

Cost of materials and of labor place a floor under the sustainable price of any item. They do not impose a ceiling.

The variables \(\displaystyle m_1, \ m_2, \ m_3\) represent weights of metals used in each item expressed in a consistent unit of weight, say ounces of metal used per item.

The functions \(\displaystyle p_1(t), \ p_2(t), \ p_3(t)\) represent units of money per unit of weight of metal, say dollars per ounce of metal.

And v(t) represents the price of one of the items. Everything then is expressed in monetary terms, and x does not represent a weight. It represents how much greater the price of the item is than the cost of the metals used in manufacturing the item.

It is about finding how much greater is the price of the item than the cost of those three materials. If that differential in value is constant in time, and only then, you can determine the weights of metal used by solving a system of four simultaneous equations.

P.S. After reading Dr. Peterson's response, if your observations are close in time and the percentage differences in the price of the item are small relative to the percentage changes in the costs of the metal, then the assumption that x is constant will be a very plausible assumption. Did you understand his comment about extra data to check the validity of the assumption?

Last edited:

- Joined
- Mar 18, 2021

- Messages
- 9

\(\displaystyle v_1 = m_1p_{1,1} + m_2p_{2,1} + m_3p_{3,1}.\)

\(\displaystyle v_2 = m_1p_{1,2} + m_2p_{2,2} + m_3p_{3,2}.\)

\(\displaystyle v_3 = m_1p_{1,3} + m_2p_{2,3} + m_3p_{3,3}.\)

Solve that system of simultaneous equations for each type of converter and check it against some other points.

- Joined
- Mar 18, 2021

- Messages
- 9

Ok I will try working that out and if it only requires 3 points and I have about 40, I should be able to use the others to check and make sure the numbers match and it is working correctly! Thank you very much.

- Joined
- Mar 18, 2021

- Messages
- 9

I guess I jumped the gun a little, I don’t fully understand the equation. The V is the value of item? And the m and p’s are metal price? What is the 1,2,and 3 at the end of the mp? Like on m1p1,1 what is the one after the comma? And where is the variable I am solving for, the x so to speak? I’ll try to do some research and figure it out. Thank you again

\(\displaystyle v_1 = m_1p_{1,1} + m_2p_{2,1} + m_3p_{3,1}.\)

\(\displaystyle v_2 = m_1p_{1,2} + m_2p_{2,2} + m_3p_{3,2}.\)

\(\displaystyle v_3 = m_1p_{1,3} + m_2p_{2,3} + m_3p_{3,3}.\)

Solve that system of simultaneous equations for each type of converter and check it against some other points.

The v’s are the value of the item at three different times. The p’s are the prices of the three different metals at times corresponding to the times when you priced the item. The m’s are your three unknowns, namely the weight of recoverable metal in a single item.I guess I jumped the gun a little, I don’t fully understand the equation. The V is the value of item? And the m and p’s are metal price? What is the 1,2,and 3 at the end of the mp? Like on m1p1,1 what is the one after the comma? And where is the variable I am solving for, the x so to speak? I’ll try to do some research and figure it out. Thank you again

So \(\displaystyle m_3 p_{3, 2}\) is the unknown weight of metal 3 recoverable times the unit price of the third metal at the second pricing time.

Do you know how to solve a system of 3 simultaneous equations for three unknowns?

Last edited:

- Joined
- Mar 18, 2021

- Messages
- 9

There are three different m’s.

- Joined
- Mar 18, 2021

- Messages
- 9

Would I be able to use Cramer's rule to solve this?

- Joined
- Mar 18, 2021

- Messages
- 9

Either way will work.