Greeting
Students;
In
today's lecture we will be discussing Math Concept of “Profit, Revenue function” which is up and running under our Blog Board link,
this class latest participation date is Apr 20th,
2015 @ 11:59 PM. No participation(s) will be accepted
after this due date. Feel free to ask questions if you have any.
Prof.
Malk AlHarbi
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Profit, Revenue function
The profit- revenue function is a
function that shows the gains made by a company in production in a mathematical
form.
It is equivalent to: Total revenue
(TR) – (TC)
Total revenue = P * Q
It is the total amount of income
gained from selling the finished products of a production process.
Total cost =Fixed cost + Variable
cost( Cost per unit * Quantity)
Profit revenue function = TR –TC
At maximum profit, the slope of the
profit function (Marginal profit) = 0
ddQ = dTR/ dQ – dTC / dQ = 0
dTR/ dQ = dTC/ dQ
But dTR / dQ = MR
And
dTC/ dQ = MC
Thus MR = MC for profit
maximization.
This is the necessary condition for
profit maximization.
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Example
Consider a company that has:
·
fixed
costs of $12000 and variable costs of
100 * q if the cost of producing
one unit is $100, thus total costs = Fixed costs + variable costs i.e. $12000 + 100q
·
Total
revenue of 180 *q if one unit is sold for $180
The profit revenue function = TR –
TC
180q – (12000 + 100q)
= 80q – 12000
Next, we find the Break- even value
which
Occurs when total revenue = total cost
Thus; 180q = 12000 + 100q
80q = 12000
Q = 150
The revenue = p* q
= 180 * 150
= 27000
Thus break even value = $27000
We can then plot a graph showing the break- even point, revenue
and the cost functions.
If 100 is the number of items produced, we come up with the
following graphical representation.
The break- even point on the graph is a point where the total
revenue = total cost (TR = TC)
The profit – revenue graph will be as follows:
For profit maximization, we
differentiate with respect to Q
Thus 80Q – 12000 = 0
dMR/dQ = 80
Thus profit is maximized at 80
units.
Profit = 80 *80 – 12000
= 16000 – 12000
= $4000
Your graphs were outstanding. Very easy to understand. Great job!
ReplyDeletesuper detailed and I agree with Mike, excellent work with the graphs.
ReplyDeleteSuper easy, excellent work especially with the graphics that you provided
ReplyDeletemlak,
ReplyDeletegood job with your lesson. the instructions were clear and thorough, although there were a few areas that were hard to read because of grammatical issues. your real world example is good.
professor little
Good job Mlak, very nice work specially with the graph.
ReplyDelete