Pricing

Mary sells T-shirts in a stall at the shopping centre. When she charges £15 per T-shirt, she does not sell anything, however she noticed that every time she reduces the selling prince by £1, sells 15 more t-shirts per hour. A T-shirt cost £4.50 to make and she is charged £130 a day for her stall. Mary opens her stall from 10am to 8pm.
a. Write the equation to calculate the maximum profit
b. Using the equation write in a table the price and profit for price ranging from £15 to £3
c. Plot a graph and draw a diagram to represent the profit
d. What is the maximum possible profit? What should be the selling price to achieve it?

Sir i very confused here.

Linear equation P=A -BQ
P = £15 while a = 0
b= change in price required to change in demanded by 1 unit the gradient of the line
MR = MC which the variable cost £4.50
MR= A-2BQ

b = 1/15 = 0.06667 = 0.1
From this p=a-bq
a= 15+0.1*15 = 16.5
P=16.5-0.1Q
MR=a-2bq
MR=MC
16.5-0.2Q = 4.5
Q=60
price function
P=16.5-0.1Q
P = 16.5-0.1*60
P=£10.5
Profit = Total sales – total cost ( FC+VC) 130+4.5 *60
Profit= 10.5*60 – 130+270
Profit= 630-400
Profit = £230

sir the answers are below

Given Q =0 when Price = 15 implies the demand intercept = 15
Q = 10 when price = 14 {Because Mary opens her store from 10 am to 8 pm 10 hrs and when the price is dropped to $14 the sale increases by 1 t-shirt per hour so 10 in 10 hours)
Demand Equation: using the following general Equation Q = a – bP
0 = a – 15b —-(1)
10 = a- 14b—–(2)
Subtract 1 from 2
10 = b
Substitute b in (1)
0 = a – 15*10
0 = a – 150
a = 150
Thus, our Demand Equation: Q= 150 – 10P
10P = 150 – Q
Inverse Demand Function: P = 15 – 0.1Q
Inverse Demand Function: P = 15 – 0.1Q
Also, the fixed Cost = 130
Cost per T-shirt = 4.5
Cost of Q T-shirt = 4.5Q = Variable Cost
Total Cost = Fixed Cost + Variable Cost
Total Cost = 130 + 4.5Q
Solution:
Profit = Total Revenue – Total COst
Total Revenue (TR)= P*Q
Substitute inverse demand function in TR
TR = (15 – 0.1Q)*Q

TR = 15Q – 0.1Q2
Profit = TR – TC = 15Q – 0.1Q2 – 130 -4.5Q
Profit = TR – TC = 10.5Q – 0.1Q2 – 130
Profit = 10.5Q – 0.1Q2 – 130
Maximum Profit:
1st order derivative of the profit function = 10.5 – 0.2Q
10.5 – 0.2Q = 0
Q = 10.5/0.2
Q = 52.5

Ans a) Profit function = 10.5Q – 0.1Q2 – 130 ; Profit maximizing function = 10.5 – 0.2Q = 0
Part b)
Price Q = 150 -10P TR = P*Q TC = 130 + 4.5Q Profit = TR – TC
15 0 0 130 -130
14 10 140 175 -35
13 20 260 220 40
12 30 360 265 95
11 40 440 310 130
10 50 500 355 145
9 60 540 400 140
8 70 560 445 115
7 80 560 490 70
6 90 540 535 5
5 100 500 580 -80
4 110 440 625 -185
3 120 360 670 -310

Part d)
From the Profit maximizing function in part a)
10.5 – 0.2Q = 0
Q = 10.5/0.2
Q = 52.5

At Q = 52.5 the firm makes maximum Profit
Substitute Q = 52.5 in the PRofit function in part a) Profit Function : 10.5Q – 0.1Q2 – 130
10.5*52.5 – 0.1*(52.5)2 -130
551.25 – 275.625 +130
Profit = 405.625
Thus the firms Maximum Profit = 405.625
The price is determined using Q = 52.5 in-demand function
Q= 150 – 10P
52.5 = 150 – 10P
10P = 150 – 52.5
10P = 97.5
P = 9.75
Thus at a price = 9.75 the firm maximizes its profit
Ans d) Maximum Profit = 405.625, Profit maximizing price = $9.75