a) We can write the cost function (in function of the units made) as the sum of the fixed cost (2000) and the variable cost (25*x):
[tex]C(x)=2000+25x[/tex]b) The revenue R(x) is:
[tex]R(x)=60x-0.01x^2[/tex]To find the value of x that maximizes R(x) we derive R(x) and equal it to 0:
[tex]\frac{dR}{dx}=60(1)-0.01(2x)=60-0.02x[/tex][tex]\begin{gathered} \frac{dR}{dx}=0 \\ 60-0.02x=0 \\ 60=0.02x \\ x=\frac{60}{0.02} \\ x=3000 \end{gathered}[/tex]We can now calculate the maximum revenue as R(3000):
[tex]\begin{gathered} R(3000)=60\cdot3000-0.01\cdot(3000)^2 \\ R(3000)=180000-0.01\cdot9000000 \\ R(3000)=180000-90000 \\ R(3000)=90000 \end{gathered}[/tex]c) The profit function P(x) can be calculated as the difference between the revenue and the cost:
[tex]\begin{gathered} P(x)=R(x)-C(x) \\ P(x)=(60x-0.01x^2)-(2000+25x) \\ P(x)=-0.01x^2+60x-25x-2000 \\ P(x)=-0.01x^2+35x-2000 \end{gathered}[/tex]In the same way as we did in b), we can calculate the number of units x that maximize the profit by deriving P(x) and making it equal to 0:
[tex]\begin{gathered} \frac{dP}{dx}=-0.01(2x)+35(1)-2000(0)=0 \\ -0.02x+35=0 \\ 35=0.02x \\ x=\frac{35}{0.02} \\ x=1750 \end{gathered}[/tex]The maximum profit can be then calculated as P(1750):
[tex]\begin{gathered} P(1750)=-0.01(1750)^2+35(1750)-2000 \\ P(1750)=-0.01\cdot3062500+61250-2000 \\ P(1750)=-30625+61250-2000 \\ P(1750)=28625 \end{gathered}[/tex]We can graph R(x) and P(x) as:
Answer:
a) C(x) = 2000 + 25x
b) x = 3000 units
R(3000) = $90000
c) x = 1750 units
P(1750) = $28625
