Respuesta :
From the problem, we have the following functions :
Cost Function :
[tex]C(x)=40x+250[/tex]Revenue Function :
[tex]R(x)=-0.5(x-80)^2+3200[/tex]1. Profit function is the difference between the revenue and cost function. This will be :
[tex]\begin{gathered} P(x)=R(x)-C(x) \\ P(x)=\lbrack-0.5(x-80)^2+3200\rbrack-(40x+250) \\ P(x)=\lbrack-0.5(x^2-160x+6400)+3200\rbrack-(40x+250) \\ P(x)=(-0.5x^2+80x-3200+3200)-(40x+250) \\ P(x)=-0.5x^2+80x-40x-250 \\ P(x)=-0.5x^2+40x-250 \end{gathered}[/tex]The profit function is :
[tex]P(x)=-0.5x^2+40x-250[/tex]2. The problem states that the maximum capacity of the company is 120 items.
So the domain is all integers from x = 0 to x = 120
3. The profit for x = 40 or 50 are :
[tex]\begin{gathered} \text{when x = 40} \\ P(40)=-0.5(40)^2+40(40)-250 \\ P(40)=550 \\ \text{when x = 50} \\ P(50)=-0.5(50)^2+40(50)-250 \\ P(50)=500 \end{gathered}[/tex]Profit when producing 40 items = 550
Profit when producing 50 items = 500
4. The factor here is the leading term which is -0.5x^2.
Take note that the sign is negative so this term will be deducted from the whole function.
The greater the value it has, the greater the deduction and will give a lesser profit.
Between 40 and 50, 50 will yield a greater deduction value. So producing 10 more units from 40 will make less profit.