Question

Find elasticity of demand and MR for the demand function at p= 70; demand function is p= 100-x-x².​

Answer 1

Answer:

Join / Login

Question

The demand for a certain product is represented by the equation p=500+25x−3x2 in rupees where x is the number of units and p is the price 3 per unit. Find:

(i) Marginal revenue function.

(ii) The marginal revenue when 10 units are sold.

Hard

Open in App

Solution

Verified by Toppr

 i )    To   find   Marginal  Revenue  function  

        Demand   for   a   certain  Product   is  represented   by  the  Equation

         p=500+25x−3x2        

        Where  x   is  the  number  units  and  p  is  the  price per  unit     

        Marginal  Revenue   function  is  the   derivative   of  the   revenue   function

        So ,  Revenue  Function  is 

       R=x.p

       R=x.(500+25x−3x2)

       R=(500x+25x2−3x3)  

 

      Now , Marginal  Revenue  function   can  be  Calculated  as 

     =dxdR

     =dxd(500x+25x2−3x3)

     =(500+50x−33x2)

     =(500+50x−x

Answer 2

Explanation:

p = 100 - x - x*2

(P=70)

70= 100- x - x*2

x = -6 or x = 5( quantity is not negetive than we use x= 5)

total revenue (tr) = Price(p) . quantity (x).

tr = (100-x-x*2) . x

= 100x- x*2-x*3

hence MR = d tr/DX= 100-2x-3x*2

put x=5

than MR =15 answer

since AR = price

than AR =100-x-x*2

put x=5

than AR =70

we know that

e =AR / AR - MR

put all the value

e = 70 / 70-15

= 70 / 55

= 1.27 Answer