Question

Que.3 The demand function for a manufacturer's product is p=(80-x))/4, where
x is the number of units and p is price per unit. At what value of x will there be
maximum revenue? What is maximum revenue?​

Answer 1

Given : The demand function for a manufacturer's product is p=(80-x) /4, where x is the number of units and p is price per unit.

To Find :   At what value of x will there be maximum revenue? What is maximum revenue?​

Solution:

Revenue = units * price

=> revenue = x *  (80-x) /4

=> revenue  =   20x  - x²/4

R(x) =  revenue function

R(x) =  20x  - x²/4

=> R'(x) = 20  - x/2

R'(x) = 0  => x = 40

R''(x)  = - 1/2 < 0

Hence maximum revenue at x = 40

revenue = 40 *  (80-40) /4    = 400

x = 40 gives maximum revenue

and maximum revenue is 400  

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