C(x)=0,01x^2-2x+25 P(x)=R(x)-C(x) R(x)=30x What is max P(x)?The company sells each unit of the product for $ 30. This is company's total cost function, where x is the number of units produced....

C(x)=0,01x^2-2x+25

P(x)=R(x)-C(x)

R(x)=30x

What is max P(x)?

The company sells each unit of the product for $ 30. This is company's total cost function, where x is the number of units produced. Determine the number of units to be produced and sold the firm to get the maximum profit. What is the maximum profit?

Expert Answers
sciencesolve eNotes educator| Certified Educator

You need to find the profit function P(x) using the information provided by problem such that:

`P(x) = R(x) - C(x)`

You need to substitute `30 x`  for `R(x)`  and `0.01x^2-2x+25`  for `C(x)`  such that:

`P(x) = 30 x - (0.01x^2-2x+25)`

You need to open brackets multiplying each term by -1 such that:

`P(x) = 30 x -0.01x^2+ 2x- 25`

You need to collect like terms such that:

`P(x) =- 0.01x^2 + 32 x - 25`

You need to optimize the profit, hence you need to differentiate the profit function with respect to x such that:

`P'(x) = - 0.02x + 32`

You need to solve the equation P'(x) = 0 such that:

`- 0.02x + 32 = 0 `

Dividing by -2 yields:

`0.01 x - 16 = 0`

`0.01 x = 16 =gt x = 16/0.01 =gt x = 1600`

The company needs to sell 1600 units of sell to get the maximum profit.

You need to substitute 1600 for x in equation of profit such that:

`P(1600) = - 0.01*1600^2 + 32*1600 - 25`

`P(1600) = - 25625 + 51200`

`P(1600) = 25575`

Hence, the company sells 1600 units and gets the maximum profit of 25575.

hala718 eNotes educator| Certified Educator

P(x) = R(x) - C(x)

==> P(x)= 30x - (0.01x^2 -2x + 25)

==> p(x)= -0.01x^2 + 32x - 25

To find the maximum number, first we will find the derivative's zero.

==> p'(x) = -0.02x + 32  = 0

==> -0.02x = -32

==> x = 32/0.02 = 1600

Then, the number of units is 1600 unit.

==> p(1600)= (0.01 1600^2) + 32(1600) -25

==> p(1600) = 25575

Then, the maximum profit is 25575