Answer:
The price that maximize profit is P = $251.5
Step-by-step explanation:
The given function for the demand, D = 2,000 - 4·P
The function that represent the total cost, C = 6,000 + 3·D
Therefore, the revenue, R = P × D = P × (2,000 - 4·P) = 2000·P - 4·P²
The cost, C = 6000 + 3·D = 6,000 + 3 × (2,000 - 4·P) = 12,000 - 12·P
The profit, Pr = R - C = 2000·P - 4·P² - (12,000 - 12·P) = 2,012·P - 12,000 - 4·P²
∴ The profit, Pr = 2,012·P - 12,000 - 4·P²
Implementing the model on MS Excel with the Solver app, by implementing the following steps;
1) Enter a value for a price into a cell on MS Excel for example A1
2) Enter the equation into another cell on MS Excel by the first cell, entering the equal to sign, '=', before the equation while replacing 'P' in the original equation with the name of the price cell as follows;
'= 2012*A1 - 12000 - 4*A1^2
3) Ensure that the Solver Add-In is enabled on MS Excel from within the Options dialogue box opened from the Files menu
4) Select the Data tab then select Solver to open the Solver dialog box
5) For the Set Objective select the cell containing the formula then for the By Changing Variable Cells box, select cell A1
6) Ensure that Max is selected in the To options then select Solve; The maximum value will be displayed in the formula cell while the price that gives the maximum value will be displayed in A1
From MS Excel Solver application, the price that gives the maximum profit is P = $251.5
