# A manufacturer offers price per unit based on, order quantity “x” as per following function: `p = 20 – 0.001x` Rs . The cost of manufacturing for X units is given by the function: `C = 5x +...

A manufacturer offers price per unit based on, order quantity “x” as per following function: `p = 20 – 0.001x` Rs . The cost of manufacturing for X units is given by the function: `C = 5x + 2000` Rs. Determine the number units he should make and sell per week to maximize the profit

*print*Print*list*Cite

First, determine the revenue (R). To do so, apply the formula:

R=(Price per unit)* (Number of units sold)

Since the price per unit is represented by the function p and x is the number of units, then:

`R=px = (20-0.001x)(x)`

`R=20x-0.001x^2`

Now that the revenue is known, apply the formula of profit which is:

Profit = Revenue - Cost

Since cost is represented by the function C, then profit (P) is:

`P= R - C`

`P= 20x- 0.001x^2 - (5x+ 2000)`

`P =20x -0.001x^2-5x-2000`

`P =-0.001x^2+15x-2000`

Now that the profit function is known, take the derivative of P to determine the value of x that would maximize the profit.

`P'=-0.001*2x +15`

`P'=-0.002x+15`

Then, set P' equal to zero.

`0=-0.002x+15`

And, isolate x.

`0+0.002x=-0.002x +15+0.002x`

`0.002x=15`

`(0.002x)/0.002x=15/0.002`

`x=7500`

**Hence, to maximize the profit, the manufacturer should produce and sell 7500 units.**