SUMMARY MATHEMATICS FOR PREMASTER - week 7-11 - endterm material

Mathematics for premaster
week 8 – 12 – endterm

Lecture 13 – week 7 (last week before midterm)

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