Respuesta :
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:
[tex]P(x)=0.5*(x-110)^2+6050-(40x+150)[/tex]Solve the equation, by solving first the quadratic part.
[tex]\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}[/tex]Then, sum the like-terms.
[tex]P(x)=-0.5x^2+70x-150[/tex]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.
[tex]\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}[/tex][tex]\begin{gathered} P(x)=-0.5x^{2}+70x-150 \\ P(80)=-0.5*80^2+70*80-150 \\ P(80)=2250 \end{gathered}[/tex]Comparing both profits, the profit for x = 70 is greater. So, they should choose x = 70.
In summary:
(a) Profit equation:
[tex]P(x)=-0.5x^{2}+70x-150[/tex](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.
