Matrix for Cost, Revenue, break-even point

[kt]

3. A cookware manufacturer is preparing to market a new pasta machine. The company’s fixed costs for research, development, tooling, and so on, are $243,000 and the variable costs are $22.45 per machine. The company sells the pasta machine for $59.95.

a) Find the cost and revenue equations. Remember to declare subscripted variables.

x: number of machines
Cost C = 243,00 + 22.45x
Revenue R = 59.95 x

b) Find the break-even point using matrices. Show all work and describe your conclusion using appropriate units of measure.

How do I answer part (b) in matrices form? I known what it is: the breakeven point is found by putting the Cost equation equal to the Revenue equation, and solving. But that is not the way he wants it, I do not think?

 

[Unco]

The company pays variable costs on the number of machines it makes, not sells, so there are two variables to consider.

 

[Denis]

Not sure what the question/problem is; ANYHOO:

if it costs $243,000 to set up shop;
and one machine nets 59.95 - 22.45 = $37.50;
then the poor guy must sell 243000 / 37.5 = 6480 machines before
he makes any profit :shock:

 

[kt]

Problem

I think he wants us to find the break even point, with a matrix, I donot known how to set it up in that form. THat is excally how he put the question on the worksheet. I did find the break even point though.

 

[steve_b]

Solving this using matrices is not the most efficient method, but here goes.

At break-even, the C will equal R, so let's call them y for simplicity.

The equations now look like:

y = 243000 + 22.45x

y = 59.95x

Rewrite the equations with the unknowns on the left and the constants on the right:

-22.45x + y = 243000

-59.95x + y = 0

The matrices are now:

[-22.45     1]
[-59.95     1]

and

[243000]
[  0   ]

The matrix equation is:

[-22.45    1]  X  [x] = [243000]
[-59.95    1]     [y]   [   0  ]

I know two ways to solve this matrix equation. One is to multiply both sides of the equation (that is, "multiply on the left") by the inverse of the coefficient matrix, and the other is to let my calculator do a "reduced row echelon" of the "augmented matrix".

[Please don't ask me to give the details of either one here :-( ]

Hope that helps...

Steve

 

[kt]

Thank you it does.