Functions Limits Differentiation Differentiation Max and Min
Profit, Cost, and Revenue Linear Programming Integration
I will try to add on more each week, so keep looking at this space.
If you find a mistake, please tell me. I may give you a surprise
Profit, Cost, and Revenue Functions
Revenue Function, R(x) Total income from producing units.
Cost Function, C(x) Total cost of producing the units.
Profit Function, P(x) Total Income minus Total Cost.
Profit = Income - Cost
P(x) = R(x) - C(x)
Marginal is rate of change of cost, revenue or profit with the respect to the number of units.
This means differentiate the cost, revenue or profit. Using any of the ways of differentiation.
Marginal Revenue, R’(x) The derivative of R(x).
Marginal Cost Function, C’(x) The derivative of C(x).
Marginal Profit Function, P’(x) The derivative of P(x).
In real words, the word “marginal” can be read as “ the next unit”, so
Marginal Revenue, R’(x) The next unit will bring this amount of Revenue.
Marginal Cost Function, C‘(x) The next unit will Cost this amount.
Marginal Profit Function, P’(x) The next unit will make a Profit of this amount.
A little hint:
Marginal and Approximate are words often used in this type of question so think MAN
Marginal or Approximate gives the Next item. Back to question
Average value is just as usual dividing the amount by the total number of items.
Average Revenue Total Revenue from n items divided by n items.
Average Cost Total Cost of n items divided by n items.
Average Profit Total Profit from n items divided by n items.
So these questions involve substituting values and differentiation.
Example One
A manufacturing company produces and sells tables. The cost function is given by
C(x) = 4x + 120Ö x + 4000 where x is the number of tables
The tables are sold for $200 each.
Find the following:
- 1. The total cost of producing 25 tables.
- 2. The total revenue and total profit from selling 25 tables.
- 3. The approximate cost of producing the 201st table.
- 4. The approximate profit on the next table after 200 tables have been sold.
- 5. The average cost per table of 200 tables.
- Don’t be worried as it really is simple, just put 25 wherever you see x in C(x).
- Revenue is Income, Cost is expense and the difference (Revenue - Cost) is Profit or Loss.
- The approximate cost of producing the 201st table.
This is how you go:
C(x) = 4x + 120Ö x + 4000.
C(25) = 4(25) + 120(Ö 25) + 4000.
C(25) = 100 + 600 + 4000 = 4700
Now you know me, I like that little sentence at the end. Just use the question to help you
write the sentence.
The total cost of producing 25 tables is $4700.
So the Revenue is the amount you sell the tables for multiplied by how many tables.
Once again put x = 25
R(x) = 200 x =200(25) = 5000
P(x) = R(x) -C(x) = 5000 - 4700 = 300 (4700 came from part 1)
Remember those sentences:
The total revenue from selling 25 tables is $5000.
The total profit from selling 25 tables is $300.
This is a little trickier than the first two parts, look back to the Marginal Cost. So we want
to know after producing 200 tables, how much does it cost to produce one more table.
Another way this could have been written is after producing 200 tables, how much does it
cost for the next table. Remember think MAN.
So read the words very carefully.
We will have to differentiate C(x) first
C(x) = 4x + 120 Ö x + 4000
C(x) = 4x + 120 x 1 / 2 + 4000
C’(x) = 4 + 120 1/2 x1 / 2 - 1 + 0
Usually this is just simple differentiation
C’(x) = 4 + 60 x - 1/2
Now the x-value that we are going to use is 200 as the next table will be 201st.
C’(x) = 4 + 60( 200 - 1/2 )
C’(x) = 4 + 60(0.0707)
C’(x) = 4 + 4.242 = 8.242
The approximate cost of producing the 201st table is $8.24.
4. The approximate profit on the next table after 200 tables have been sold.
R(x) = 200x C(x) = 4x + 120Ö x + 4000
P(x) = R(x) - C(x) = 200x - (4x + 120Ö x + 4000) = 196x - 120Ö x - 4000
Now we see the word approximate and we think differentiate and put 200 in for the value
of x. Remember 201 would be the next number after 200.
P’(x) = 196 - 60 x - 1/2
When x = 200, P’(x) = 196 -60 ( 200 - 1/2 ) = 196 - 60 (0.0707)
P’(x) = 196 - 4.242 = 191.758
The approximate profit on the next table after 200 tables have been sold is $191.76.
5. The average cost per table of 200 tables.
This is just like any other average so find the cost for 200 tables and then divide by 200.
C(x) = 4x + 120 Ö x + 4000.
The cost of 200 tables is
C(200) = 4(200) + 120Ö 200 + 4000.
C(200) = 800 + 120(14.1421) + 4000 = $6497.06
The average cost is dividing this amount by 200.
Average Cost (200) = 6497.06 / 200 = $32.49
The average cost per table of 200 tables is $32.49.
This again is not so difficult as you first think but it does require you to think! We will work
through an example as it is easier that way.
Example One
A farmer has 100 acres on which he can plant two crops; wheat or barley.
The following table gives the expenses associated with each crop.
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The farmer has 4,000 m3 of storage space. Each acre yields an average of 110 m3 of Barley and
30m3 Wheat. The farmer has available capital of $15 000.
If the net profit per m3 of Barley is $1.30 and for Wheat is $2.00, this is after all expenses have
been subtracted. How should the farmer plant the 100 acres to maximise the profits?
The first step is to say what the letters are standing for and to do this look at the question.
The last sentence is “How should the farmer plant the 100 acres to maximise the profits?” helps us
as we see that we need to know how many acres of each crop to plant.
Let x be the number of acres of Barley planted.
Let y be the number of acres of Wheat planted.
Sometimes what the letters are standing for may be given to you.
Now let us look at the question, we are given some information about how many acres,
how much money and how much storage the farmer has. So in these types of questions, look
for what you are told about the situation as a fact. Another hint is put all of the quantities
with the same units together i.e. m3 , acres and $.
“A farmer has 100 acres on which he can plant two crops...” so he can not plant more than 100
acres but sometimes it could be profitable for him to plant less than 100 acres.
Now we have to write that as a mathematical statement:
x + y £ 100
From the question we know that “The farmer has available capital of $15 000.” so his expenses
cannot be more than this value. So we must work out what the expense is for planting each of
the crops.
Look back at the table, to see the cost for Barley and Wheat, the numbers in Green.
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If x is the number of acres of Barley planted, then the cost of planting Barley is $120x.
If y is the number of acres of Wheat planted. then the cost of planting Wheat is $210y.
Putting all of that together, the farmer cannot spend more than $15 000 and his cost for the crops
is (120x plus 210y). In mathematical terms,
120x + 210y £ 15 000
From the question we are told that “The farmer has 4,000 m3 of storage space” so how much
room does Barley and Wheat take? The question tells us “Each acre yields an average of 110 m3
of Barley
and 30m3 Wheat”. So writing this mathematically, we have
110x + 30y £ 4 000
Now we want to know the maximum profit for the farmer. So let us think, what do we have
to do to find the profit for each crop. In the question, we are told the profit per m3
“profit per m3 of Barley is $1.30 and for Wheat is $2.00 “and the average yield for each crop
“Each acre yields an average of 110 m3 of Barley and 30m3 of Wheat” so multiplying these two quantities together for each crop and then adding the two will give the overall profit for both
crops. So writing this mathematically, we have
Profit = (1.30) 110x + (2.00) 30y
Profit = 143x + 60y
So let us now collect all the inequalities together
x + y £ 100
120x + 210y £ 15 000
110x + 30y £ 4 000
Profit = 143x + 60y, this is what we want to maximise
There is another way of producing the same inequalities by using the following diagram
and filling in the spaces
Line One
“A farmer has 100 acres on which he can plant two crops...” so he can not plant more than 100
acres but sometimes it could be profitable for him to plant less than 100 acres.
We fill the pink ovals with how many acres of Barley and Wheat and that is x and y respectively
and put the amount of acres available at the end.
Line Two
From the question we know that “The farmer has available capital of $15 000.” so his expenses
cannot be more than this value. So we must work out what the expense is for planting each of
the crops. The cost of planting Barley is $120x and the cost of planting Wheat is $210y.
We fill the blue rectangles with how much it costs to plant Barley and Wheat and that is $120
per acre for Barley and $210 per acre for Wheat and put the amount of money available at the end.
Line Three
From the question we are told that “The farmer has 4,000 m3 of storage space” so how much room
does Barley and Wheat take? The question tells us “Each acre yields an average of 110 m3 of
Barley and 30m3 Wheat”.
We fill the green circles with how much storage Barley and Wheat take and that is 110m3 and
30m3 respectively and put the amount of storage available at the end.
Line Four
In the question, we are told the profit per m3,“profit per m3 of Barley is $1.30 and for Wheat is
$2.00” and the average yield for each crop “Each acre yields an average of 110 m3 of Barley and
30m3 of Wheat” so multiplying these two quantities together for each crop and then adding the
two will give the overall profit for both crops.
We fill the brown rectangles with how much profit made by Barley and Wheat, $143 and $60 resp.
This is what we want to maximise so there is no number at the end
So let us now write all the inequalities together
x + y £ 100
120x + 210y £ 15 000
110x + 30y £ 4 000
Profit = 143x + 60y, this is what we want to maximise
We have the same inequalities as before just a more picturesque way of finding them.
Now this is where we will get serious and have to graph all these lines on a diagram at one time.
Now this is where my diagrams may not look so good due to my lack of computing skills but please
try to follow it.
How do we draw the inequalities as lines, we will think of the equations as equal to rather than
less than or greater than, then we will put x = 0 and find out the y-value and then put y = 0 and
find out the x-value. These are the y and x intercepts on the graph.
Let’s show how we can do that using 110x + 30y = 4 000
Let x = 0 so 30y = 4 000 y = 400/3 = 133.333.
Let y = 0 so 110x = 4 000 x = 400/11 = 36.3636
Once we have shaded the correct side to the left for less than and to the right for greater than
then we will mark the corners of that area.
The pink line is 110x + 30y = 4 000
The green line is x + y = 100
The blue line is 120x + 210y = 15 000
The area required is shaded in yellow
The multicoloured dots A, B, C and D are the points of interest as these are the extremes of the
shaded area.
We will have to find the co-ordinates of these points, some are easy and others are not
Point A x = 0, 120x + 210y = 15 000, 210y = 15 000 y = 500/7 = 71.429
Point B x = 0 y = 0
Point C y = 0 110x + 30y = 4 000, 110x = 4 000 x = 400/11 = 36.364
Point D needs a bit of work as it is not on any of the axes.
We will have to solve either on a graphics calculator or simultaneous equations to find the intersection point.
110x + 30y = 4 000 1 multiply each term by 7 770x + 210y = 28 000 3
120x + 210y = 15 000 2
770x + 210y = 28 000 3 Equation 3 minus Equation 2
120x + 210y = 15 000 2
650x = 13 000 so x = 13 000/650 = 20
Substituting the value of x into equation 1 (any of the equations would do)
110(20) + 30y = 4 000
2200 + 30y = 4 000
30y = 1 800 so y = 1 800/30 = 60.
So the co-ordinates of point D are x = 20 and y = 60.
We find the following table by writing in the co-ordinates and then the formula which we want
to maximise or minimise. If we want to maximise then we choose the largest value and the
opposite if we want to minimise i.e. the smallest.
So we can see that the formula, Profit = 143x + 60y is a maximum at point D,
where x = 20 and y = 60 as seen in the table below.
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The value of the maximum profit is $ 6 460 and occurs by planting 20 acres of Barley and 60 acres
of Wheat
So with a few variations this is a template to follow for Graphical Solutions or Linear
Programming
