SUMMARY MATHEMATICS FOR PREMASTER - week 7-11 - endterm material
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Course
Mathematics
Institution
Tilburg University (UVT)
Detailed summary of all lecture material including examples and exercises, focused on the material discussed in weeks 7 - 11. All material for the endterm of mathematics for premaster
Supply curve – optimisation function of 1 variable
1) Profit maximisation
Revenue function = R(y) = 8y y>0
3 2
Cost function = C(y) = y – 4y + 8y y > 0 (domain restriction)
Aim is to maximise the profit function = ╥(y) , with y ≥ 0
o Profit function = revenue – cost
╥(y) = R(y) – C(y)
╥(y) = 8y - y3 + 4y2 + 8y
= -y3 + 4y2
o Stationary points = get the derivative of the revenue & cost function, put the
function equal to 0 to retrieve the stationary points
R’(y) = -3y2 + 8y = 0
Y(-3y + 8) = 0 -> y=0 or -3y + 8 = 0, y= 8/3
Make sing chart (remember the domain, boundary) put in the stationary
points and see if it is positive or negative on the intervals
Plug the stationary point in to the profit function and find the maximum
point = this is the point where the profit is maximal (highest point is reached,
but you do not know if the profit is positive)
Use MR(y) = MC(y) to check the stationary point
The derivative of a revenue function = the marginal revenue function = MR(y). this is the same for a
cost function, the derivative is the marginal cost function
R’(y) = MR(y)
C’(y) = MC(y)
MR(y) = MC(y) -> must be equal to each other
Marginal output rule = the output quantity y > 0 (you look at an interior point) that maximises the
profit, ╥(y) = R(y) – C(y), satisfies the equation that MR(y) = MC(y)
Production rule = if y > 0 is the output quantity that maximises profit, the producer will produce if;
AR(y) ≥ AC(y)
The AR per unit is larger than the AC per unit, so you make a profit -> revenue larger than the
costs
o ╥(y) = R(y) – C(y)
o Y *( r(y) / y – C(y) / y)
o Y *(AR(y) – CR(y) ) -> both sides are > 0
- AR(y) = average revenue at production y
o AR(y) = R(y) / y
- AC(y) = average cost at production y
o AC(y) = C(y) / y
, Example – supply function; price-taking producer
Revenue function = R(y) = py , y>0, P>0, where p indicates the price of the product
Cost function = C(y) = y3 – 4y2 + 8y y >0
Determine for each price (p) the output quantity that maximises profit
1) AR(y) ≥ AC(y)
a. AR(y) = R(y) / y = Py / y = p
Need to check where P ≥ AC(y)
AC(y) = C(y) / y = y2 – 4y + 8
o A’C(y) = 2y-4 <-> AC’(y) = 0, if y =2
o Make sing chart to determine the maximum and minimum locations
minimum location = AC(2) = 4
Apply production rule, if p= 4, then positive profits are possible, when p <4, stop producing
because no positive profits to be made
Now determine the level of the profits, what is the maximal profit level. Use the marginal
revenue
o MR(y) = MC(y)
o P = MC(y) <-> y = MC’(p)
o Need to solve that p = equal to the MC
P = 3y2 -8y +8 (recognise the quadratic equation, ABC formula)
get everything to the right, to put equal to 0
3y2 -8y +8 – p = 0
A = 3, b= -8, c= 8-p
Y = -(-8) +- √(64 – 4*3*(8-p) / 2*3 = 8/6 +- 1/6 √(64 – 12*(8-p)
Make sing chart
Y(p) = 0, when p<4
Y(p) = 8/6 + 1/6 √(64 – 12*(8-p), when p ≥ 4
This is the y that maximises the profit
Supply function
Supply function of a producer with profit function ╥(y) = py – C(y) is given by;
Y(p) = MC-1(p) when P ≥ minimum value of the AC(y)
0 when p < minimum value of the AC(y)
For 0 ≤ p < minimum AC(y),the producer does not make any profit. The maximal profit is
negative & so not profit
Only for interior points
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