Advertisements
Advertisements
Question
For the demand function p x = 100 - 6x2, find the marginal revenue and also show that MR = p`[1 - 1/eta_"d"]`
Advertisements
Solution
Given p = 100 - 6x2
we know that R = px
R = `(100 - 6x^2)x = 100x - 6x^3`
Marginal Revenue (MR) = `"dR"/"dx" = "d"/"dx" (100x - 6x^3)`
LHS = MR = `100 - 18x^2` ....(1)
Differentiating p, with respect to, 'x' we get,
`"dp"/"dx" = - 12 x`
`"dx"/"dp" = (-1)/(12x)`
`therefore eta_"d" = - "p"/x * "dx"/"dp"`
`= (- (100 - 6x^2))/x * ((-1)/(12x))`
`= (100 - 6x^2)/(12x^2)`
`therefore "RHS" = "p"(1-1/(eta_"d"))`
`= (100 - 6x^2)(1 - (12x^2)/(100 - 6x^2))`
RHS = `(cancel(100 - 6x^2))((100 - 6x^2 - 12x^2)/cancel(100 - 6x^2))`
= 100 - 18x2 ...(2)
From (1) and (2), LHS = RHS
`therefore "MR" = "p"(1 - 1/eta_"d")`
APPEARS IN
RELATED QUESTIONS
The total cost of x units of output of a firm is given by C = `2/3x + 35/2`. Find the
- cost when output is 4 units
- average cost when output is 10 units
- marginal cost when output is 3 units
The supply function of certain goods is given by x = a`sqrt("p" - "b")` where p is unit price, a and b are constants with p > b. Find elasticity of supply at p = 2b.
Show that MR = p`[1 - 1/eta_"d"]` for the demand function p = 400 – 2x – 3x2 where p is unit price and x is quantity demand.
For the demand function p = 550 – 3x – 6x2 where x is quantity demand and p is unit price. Show that MR =
For the demand function x = `25/"p"^4`, 1 ≤ p ≤ 5, determine the elasticity of demand.
Find the equilibrium price and equilibrium quantity for the following functions.
Demand: x = 100 – 2p and supply: x = 3p – 50.
The demand and cost functions of a firm are x = 6000 – 30p and C = 72000 + 60x respectively. Find the level of output and price at which the profit is maximum.
Find out the indicated elasticity for the following function:
p = xex, x > 0; ηs
Find the elasticity of supply when the supply function is given by x = 2p2 + 5 at p = 1.
A company begins to earn profit at:
