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...

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