Maximizing Revenue

[nycmath]

Precalculus
Michael Sullivan
Edition 10
Chapter 3, Section 3.4

See attachment

[Eigenvalue]

a) R=px=p(-5p+100)=-5p²+100p

b) Let p=15
R=-5(15)²+100(15)=-5(225)+1500=375
The revenue is 375 dollars

c)To find the maximum revenue, find the vertex
p=[tex]\frac{-b}{2a}[/tex]=[tex]\frac{-100}{2(-5)}[/tex]=10
x=-5p+100=-5(10)+100=50
R=10(-5(10)+100)=10*50=500
The quantity 50 objects would maximize revenue; the maximum revenue is 500 dollars
d) The company should charge 10 dollars to maximize revenue

e) Let R=480
-5p²+100p[tex]\ge[/tex]480
-5p²+100p-480[tex]\ge[/tex]0
-5(p²-20p+96)[tex]\ge[/tex]0
p²-20p+96[tex]\le[/tex]0
(p-8)(p-12)[tex]\le[/tex]0
8[tex]\le[/tex]p[tex]\le[/tex]12

The company should charge between 8 to 12 dollars to generate at least 480 dollars in revenue