If the demand function `p` is given by `x=(600-p)/8`, where the price is Rs `p` per unit and the manufacturer produces `x` unit per week at the total cost of Rs. `x^2+78x+2500`, then find the value of `x` for which the profit is maximum

If the demand function is p=200-4x , where x is the number of units demand and p is the price per unit, the marginal revenue is

If the demand function is p=200-4x , where x is the number of units demanded and p is the price per unit, then MR is

If the demand function is x= (24-2p)/(3) where x is the number of units produced and p is the price per unit, then the revenue function R(x) is

The demand function is x=(24-2p)/(3) where x is the number of units demanded and p is the price per unit. Find : (i) The revenue function R in terms of p. (ii) The price and the number of units demanded for which the revenue is maximum.

The demand function is x = (24 - 2p)/(3) , where x is the number of units demanded and p is the price per unit. Find the revenue function R in terms of p.

The demand function of a certain commodity is given by p= 1000 - 25 x+x^(2) where 0le x le20 . Find the price per unit and total revenue from the sale of 5 units.

If the demand function for a product is p=(80-pi)/(4) , where x is the number of units and p is the price per unit, the value of x for which the revenue will be maximum is

A radio manufacturer produces x sets per week at a total cost of Rs. (x^(2)+78x+2500) . . He is a monopolist and the demand function for his product is x=(600-p)/(8) , where the price is Rs. P per set . Show that the maximum net revenue (i.e., profit is obtained when 29 sets are produced per week What is the monopoly price ?

A company is selling a certain product. The demand function of the product is linear. The company can sell 2000 units, when the price is Rs8 per unit and when the price is Rs4 per unit, it can sell 3000 units. Determine: (i) the demand function (ii) the total revenue function.