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- February 18, 2024 at 12:17 pm #700595
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-2BQb = 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 = £230sir 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)*QTR = 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 = 52.5Ans 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 -310Part d)

From the Profit maximizing function in part a)

10.5 – 0.2Q = 0

Q = 10.5/0.2

Q = 52.5At 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 = $9.75February 18, 2024 at 1:16 pm #700600Where is this question from?

So exactly what do you want me to doYou have the answer to this question

Work through the answer and see where you have gone wrong.February 19, 2024 at 8:38 pm #700691This is why i do not post the work book answers to avoid overload inforamtion.

My question is , the answers are very different from my mine as you can see from the lenghty workings

I simply adopted your video steps and Kaplan but it appears the results are not same.The profit i got was £230 while the book got over £405.625

February 19, 2024 at 11:15 pm #700700I don’t believe this is a past ACCA question or is it taken from the BPP or Kaplan kit…is it?

Your calculation of a and b is wrong

As you can see for yourself

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 = bSubstitute b in (1)

0 = a – 15*10

0 = a – 150

a = 150Thus, Demand Equation: Q= 150 – 10P

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 + 4.5QEtc

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