Advertisements
Advertisements
Question
The total cost function y for x units is given by y = 3x`((x+7)/(x+5)) + 5`. Show that the marginal cost decreases continuously as the output increases.
Advertisements
Solution
The total cost function, y = `3x((x+7)/(x+5)) + 5`
To prove the marginal cost decreases continuously as the output increase we should prove `"dy"/"dx"` is positive.
y = `3x((x+7)/(x+5)) + 5`
`= 3x (((x + 5) + 2)/(x + 5)) + 5`
`= 3x ((x + 5)/(x + 5) + 2/(x + 5)) + 5`
y = `3x(1 + 2/(x+ 5)) + 5`
y = `3 (x + (2x)/(x + 5)) + 5`
`"dy"/"dx" = 3 "d"/"dx" [x + (2x)/(x + 5)] + "d"/"dx" (5)`
`= 3 [1 + 2 "d"/"dx" (x/(x + 5))] + 0`
`= 3 [1 + 2(((x + 5)1 - x(1))/(x+5)^2)]`
`= 3 [1 + 2((x + 5 - x)/(x+5)^2)]`
`= 3 [1 + 2(5/(x + 5)^2)]`
`= 3 [1 + 10/(x+5)^2]`, which is positive.
∴ The marginal cost decreases continuously of the output increases.
APPEARS IN
RELATED QUESTIONS
A firm produces x tonnes of output at a total cost of C(x) = `1/10x^3 - 4x^2 - 20x + 7` find the
- average cost
- average variable cost
- average fixed cost
- marginal cost and
- marginal average cost.
The demand curve of a commodity is given by p = `(50 - x)/5`, find the marginal revenue for any output x and also find marginal revenue at x = 0 and x = 25?
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.
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.
The cost function of a firm is C = x3 – 12x2 + 48x. Find the level of output (x > 0) at which average cost is minimum.
Find the elasticity of supply when the supply function is given by x = 2p2 + 5 at p = 1.
For the demand function p x = 100 - 6x2, find the marginal revenue and also show that MR = p`[1 - 1/eta_"d"]`
If demand and the cost function of a firm are p = 2 – x and C = -2x2 + 2x + 7 then its profit function is:
If the demand function is said to be inelastic, then:
The elasticity of demand for the demand function x = `1/"p"` is:
