Question

The demand function of a commodity is p = `200 - x/100` and its cost is C = 40x + 120 where p is a unit price in rupees and x is the number of units produced and sold. Determine

1. profit function
2. average profit at an output of 10 units
3. marginal profit at an output of 10 units and
4. marginal average profit at an output of 10 units.

Answer 1

The demand function, p = `200 - x/100`

Cost is C = 40x + 120

Revenue function, R(x) = px

`= (200 - x/100)x`

`= 200x - x^2/100`

(i) Profit function = R(x) – C(x)

`= 200x - x^2/100 - (40x + 120)`

`= 200x - x^2/100 - 40x - 120`

`= 160x - x^2/100 - 120`

(ii) Average profit (AP) = `"Total profit"/"Output"`

`= 1/x (160x - x^2/100 - 120)`

`= 160 - x/100 - 120/x`

Average profit at an output of 10 units

When x = 10, average profit = `160 - 10/100 - 120/10`

`= 160 -1/10 - 12`

`= 148 - 1/10`

= 148 – 0.1

= ₹ 147.9

(iii) Marginal profit [MP] = `"dP"/x`

`= "d"/"dx" (160x - x^2/100 - 120)`

`= 160 - "2x"/100`

`= 160 - x/50`

Marginal profit when x = 10, is = `160 - 10/50`

`= 160 - 1/5`

= 160 – 0.2

= ₹ 159.8

(iv) Average profit AP = `160 -x/100 - 120/x`

Marginal average profit (MAP) = `"d"/"dx"`(AP)

`= "d"/"dx" (160 - x/100 - 120/x)`

`= 0 - 1/100 - 120(- 1/x^2)  ....[because "d"/"dx" (1/x) = (-1)/x^2]`

`= (-1)/100 + 120/x^2`

Marginal average profit (MAP) = `- 1/100 + 120/10^2`

=`- 1/100 + 120/100`

`= (- 1 + 120)/100`

`= 119/100`

= ₹ 1.19