Question

For a certain company, the cost function for producing x items is C(x)=40x+150 and the revenue function for selling x items is R(x)=−0.5(x−110)^2+6,050. The maximum capacity of the company is 150 items.Assuming that the company sells all that it produces, what is the profit function?P(x)=What is the domain of P(x)?The company can choose to produce either 70 or 80 items. What is their profit for each case, and which level of production should they choose?

Answer 1

To solve this question, follow the steps below.

Step 01: Find the profit function P(x).

Given

C(x) = cost of producing x units

R(x) = revenue when producing x units

Then, P(x) is = R(x) - C(x).

Substituting the equations in the formula:

`P(x)=0.5*(x-110)^2+6050-(40x+150)`

Solve the equation, by solving first the quadratic part.

`\begin{gathered} P(x)=-0.5*(x^2-2*110*x+110^2)+6050-40x-150 \\ P(x)=-0.5x^2+110x-6050+6050-40x-150 \\ P(x)=-0.5x^2+110x-40x-150 \end{gathered}`

Then, sum the like-terms.

`P(x)=-0.5x^2+70x-150`

Step 02: Find the domain.

Since the maximum capacity of the company is 150 items. So, x-maximum is 150.

The minimum number of products is 0.

Then, 0 ≤ x ≤ 150.

Domain: 0 ≤ x ≤ 150 or [0, 150].

Step 03: Compare the profit for x = 70 and x = 80.

`\begin{gathered} P(x)=-0.5x^(2)+70x-150 \\ P(70)=-0.5*70^2+70*70-150 \\ P(70)=-2450+4900-150 \\ P(70)=2300 \end{gathered}`
`\begin{gathered} P(x)=-0.5x^(2)+70x-150 \\ P(80)=-0.5*80^2+70*80-150 \\ P(80)=2250 \end{gathered}`

Comparing both profits, the profit for x = 70 is greater. So, they should choose x = 70.

In summary:

(a) Profit equation:

`P(x)=-0.5x^(2)+70x-150`

(b )Domain:

0 ≤ x ≤ 150 or [0, 150].

(c) Comparing x = 70 and x = 80.

Comparing both, the company should choose x = 70, because the profit is greater.