# Determine the optimal selling price to maximize profit in the following case:A man can sell 150 tables/month for $200 each. If he reduces the price, the number of tables sold increases by 25 for a...

Determine the optimal selling price to maximize profit in the following case:

A man can sell 150 tables/month for $200 each. If he reduces the price, the number of tables sold increases by 25 for a decrease in price by $1. The cost of manufacturing one table is $125.

### 2 Answers | Add Yours

The man can sell 150 tables a month if they are priced at $200 each. For every $1 decrease in the price, the number of tables sold increases by 25. If the price of a table is P less than $200, the number of tables sold is 150 + 25P. The revenue earned is [150 + 25P]*(200 - P)

The cost of manufacturing [150 + 25P] tables is [150 + 25P]*125

From the revenue and the cost, we get the profit as:

[150 + 25P](200 - P) - [150 + 25P]*125

=> (150 + 25P)(75 - P)

=> 11250 + 1725P - 25P^2

Differentiate 11250 + 1725P - 25P^2 with respect to P and solve the derivative for P.

1725 - 50P = 0

=> P = 1725/50 = 34.5

P is the value by which the price is less than 200, the actual price is 200 - 34.5 = 165.5

**The profit is maximized when the price of the table is $165.5**

P = a - bQ **Calculate b**

Eq 1 : 200 = a - 150b

Eq 2 : 199 = a - 175b

Eq 1 - Eq 2 = 1 = 0 - 25

**b = 1/25 ie 0.04**

** Calculate a**

200= a-(0.04x150)

200 = a - 6

**a = 206**

**Optimal price equation P = 206 - 0.04Q**

**Profit maximised where MC = MR**

MR = a-2bQ

MR = 206 - 0.08Q = 125 (MC)

= (206-125=) 81 / 0.08 = 1012.50 units

**Calculate Selling Price pu at which profit is maximised**

P = 206 - (0.04 x 1012.5)

P = 206 - 40.5

P = $165.5