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