Part A:
Given that Box office revenue at a multiplex cinema in paris is [tex]r(p) = 3600p - 6p^3[/tex] euros per showing when the ticket price is p euros.
When p = 9,
[tex]r(9) = 3600(9) - 6(9)^3 \\ \\ =32,400-6(729)=32,400-4,374 \\ \\ =\bold{28,026 \ euros}[/tex]
Part B:
The linear approximation of the change in a function Δf(x) using a value, a, close to x is given by:
[tex]L(a+Δx)=(Δx)f'(a)[/tex]
Given that [tex]r(p) = 3600p - 6p^3[/tex], then [tex]r'(p) = 3600 - 12p^2[/tex]
Using a = 9, we have:
[tex]r'(9)=3600-12(9)^2 \\ \\ =3600-12(81)=3600-972 \\ \\ =2,628[/tex]
Thus, If p is raised by 0.5 euros, then
[tex]\Delta R=(\Delta p)f'(9) \\ \\ =0.5(2,628)=\bold{1,314 \ euros} [/tex]
Part C:
The linear approximation of the change in a function Δf(x) using a value, a, close to x is given by:
[tex]L(a+Δx)=(Δx)f'(a)[/tex]
Given that [tex]r(p) = 3600p - 6p^3[/tex], then [tex]r'(p) = 3600 - 12p^2[/tex]
Using a = 9, we have:
[tex]r'(9)=3600-12(9)^2 \\ \\ =3600-12(81)=3600-972 \\ \\ =2,628[/tex]
Thus, If p is lowered by 0.5 euros, then
[tex]\Delta R=(\Delta p)f'(9) \\ \\ =-0.5(2,628)=\bold{-1,314 \ euros} [/tex]