Maximizing Revenue

[nycmath]

Precalculus
Michael Sullivan
Edition 10
Chapter 3, Section 3.4

See attachment

[Eigenvalue]

a) R=px
=x(-[tex]\frac{1}{6}[/tex]x+100)
=-[tex]\frac{1}{6}[/tex]x²+100x

b) The domain of R is all rel numbers, as the domain for a quadratic polynomial is all real numbers--there are no values for x that would evaluate to a y-value with a denominator of zero or a negative number radicand---revenue can be negative

c) Let R=200
200=[tex]\frac{-1}{6}[/tex]x²+100x
-x²+600x-1600=0
x²-600x+1600=0
(x-800)(x+200)=0
x=800, as the quantity of items sold cannot be negative


d) To maximize revenue, find the vertex
x=[tex]\frac{-b}{2a}[/tex]=[tex]\frac{-100}{-2(1/6)}[/tex]=300
R=[tex]\frac{-1}{6}[/tex](300)²+100(300)=-15000+30000=15000
The quantity 300 objects maximizes revenue, generating a revenue of 15000 dollars

e) p=[tex]\frac{-1}{6}[/tex](300)+100
=-50+100
=50
A price of 50 dollars maximizes revenue