# 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?

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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**

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.**