Profit Maximisation

June 08 Q4 is called “The Childrens Toy Company (CTC)”.

Not sure what question you are referring to.

This is a pricing question and the rule for profit maximisation is to find where marginal cost (MC) = marginal revenue (MR).

Using a profit-maximising algebraic approach where P = a – bQ [this is the standard starting point linking quantity of rooms sold and the price, P]

Then 400 = P – 0•25 x 1,440

Where a = 10/40 = 0•25 (a change of 40 people for a change of price of $10) and double occupancy = 75% x 2,400 = 1,800 x 80% = 1,440

Hence P = 400 + (0•25 x 1,440) = $760

Therefore P = 760 – 0•25Q

Total revenue = P x Q, so

TR = 760Q – 0•25Q2

MR = 760 – 0•5Q [MR = marginal revenue; it seems to me that this step, from TR to MR is beyond your syllabus now, so Don’t worry about it]

MC = 200 (2 persons x $100).

In order to achieve maximum profit then MR = MC

Therefore 760 – 0•5Q = 200
And Q = (760 – 200)/0•5 = 1,120 units

Therefore the selling price per double room at the profit-maximising level (P):
= 760 – 0•25 x 1,120
= 760 – 280
= $480

hey my name is mishaal.i have a small question about this profit max approach p = a – bx..
.i know p is the price at which profit will be maximised .x is quantity demanded..b is change in price with change in demand..i have doubt about ‘a’ can u plz tell???

@ mishaaljavaid, a would be the price at which nothing would be demanded (because it is so high).