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    Differentiation   
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. 1. The total cost of producing 25 tables.
2. 2. The total revenue and total profit from selling 25 tables.
3. 3. The approximate cost of producing the 201st table.
4. 4. The approximate profit on the next table after 200 tables have been  sold.
5. 5. The average cost per table of 200 tables.

This is how you  go:

1. Don’t be worried as it really is simple, just put 25 wherever you see x  in C(x).

C(x)  =  4x +    120Öx + 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.

2. Revenue is Income, Cost is expense and the difference (Revenue - Cost)  is Profit or Loss.

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.

3. The approximate cost of producing the 201st table.

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.
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.

Back to  top

Linear Programming

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.

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.

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  m^3
“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. 

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

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