So in this scenario the firstly when take the case 1 then the Total Revenue

# So in this scenario the firstly when take the case 1

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So, in this scenario the firstly when take the case 1 then the Total Revenue decrease as compared to case 2 Total Revenue value. Case 2 is example of Total Maximum Revenue after that point of Price increase then its total revenue Decreases.
Question4: What is optimum price at which total revenue can be maximized for Maa mustard oil if the competitors’ prices do not increase in october2015 (scenario 1)? If competitors increase their price by around 6% as suggested in case (scenario 2), what would be the optimum price? Does the company benefit if competitors? increase their price? perform all calculation under the assumption of no increase in promotional expenditure in the next month and 1% increase in the per capita income of consumer? Answer4): - Need to the find optimum Price at which Total revenue (TR) is maximized. Qmaa= 5024.58-136.62 Op + 117.40 Cp - 0.2823 IPCp + 7.87 PEp. Scenario 1): - If the competitors’ prices do not increase in october2015 (Keep Cp as constant) Qmaa= 5024.58-136.62 Op + 117.40 Cp - 0.2823 IPCp + 7.87 PEp. -----------(i) Cp = 109.14 Op = P IPCp = 7545.15 (As per Question it increases by 1%) So, IPCpn = 7620.6015 PEp = 1247.31 Qmaa = Quantity demand Putting all values of variables in Equation (i) Qmaa = 5024.58 136.62P + 117.40*109.14 0.2823*7620.6015 + 7.87 * 1247.31 Qmaa = 5024.58 - 136.62P + 12813.036 - 2151.29580345 + 9816.3297 Qmaa = 25502.64 136.62P -------------------------(ii) For Total Revenue (TR) TR = Qmaa * P Putting the value Qmaa in the above equation TR = 25502.64P 136.62P 2 ---------------------------------(iii) Taking differentiation from the sides w.r.s.t P d(TR)/d(P) = 25502.64 273.24 P As we know that Total Revenue is maximum when slope of d(TR)/d(P) equals to zero So, d(TR)/d(P) =0
0 = 25502.64 273.24 P P = 93.334 Putting the value of P in Eq (iii) we get the Total Revenue TR = 25502.64*93.334 136.62*93.334 * 93.334 TR = 2,380,263.40176 - 1,190,129.00166072 TR = 1190134.401 So, the total Revenue TR is 1190134.401 is maximum when optimum Price P is 93.334 Scenario 2) If the competitors’ prices increase in october2015 by 6 % Qmaa= 5024.58-136.62 Op + 117.40 Cp - 0.2823 IPCp + 7.87 PEp. ---------(i) Cp = 109.14 * 6 % Cp = 115.6884 Op = P IPCp = 7545.15 (As per Question it increases by 1%) So, IPCpn = 7620.6015 PEp = 1247.31 Qmaa = Quantity demand Putting all values of variables in Equation (i) Qmaa = 5024.58 136.62P + 117.40*115.6884 0.2823*7620.6015 + 7.87 * 1247.31 Qmaa = 26269.08405655 136.62P -------------------------- (ii) For Total Revenue (TR) TR = Qmaa * P Putting the value Qmaa in the above equation TR = 26269.08405655P 136.62P 2 -----------------------------(iii) Taking differentiation from the sides w.r.s.t P d(TR)/d(P) = 26269.084 273.24P As we know that Total Revenue is maximum when slope of d(TR)/d(P) equals to zero
So, d(TR)/d(P) =0 26269.084 273.24P = 0 P = 96.139

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• Summer '17