Maximum Revenue

Section 2.1
Question 72

For part (a), I must find R(4), R(6), and R(8).

Correct?

I need help for part (b).

What is maximum revenue?

 

b. write equation in vertex form , vertex is a maximum

 

Will work on this when time allows. Very busy as I get closer to the moving date which has been changed from 9/9 to 9/5.

 

I worked out part (a) on paper.

R(x) = revenues earned per day.

R(4) = $408

R(6) = $468

R(8) = $432

Part (b)

R(p) = -12p^2 + 150p

R(p) = -(12p^2 - 150p)

R(p) = -(p^2 - (150/12)p)

R(p) = -(p^2 - 150/12p + 625/16 -625/16) + 0

R(p) = -(p^2 - 150/12p + 625/16) - (-625/16) + 0

R(p) = -(p^2 - 150/12p + 625/16) + 625/16

R(p) = -(p - 150/12)(p - 150/12) + 625/16

R(p) = -(p - 150/12)^2 + 625/16

Vertex = (h, k)

Vertex = (-150/12, 625/16)

Now k = 625/16 = 39.0625

The unit price that yields a maximum revenue is
$39.0625 or just $39.

You say?

 

correction

R(p) = -12p^2 + 150p

R(p) = -(12p^2 - 150p)....factor out -12

R(p) = -12(p^2 - (150/12)p).......simplify

R(p) = -12(p^2 - (25/2)p+b^2)-12(-b^2).........b=(25/2)/2=(25/4)

R(p) = -12(p^2 - (25/2)p+(25/4)^2)+12(25/4)^2

R(p) = -12(p - 25/4)^2+1875/4

vertex: (h,k)=(p,R(p) ) is at

(25/4,1875/4) or (6.25, 468.75)

The unit price (p) that yields a maximum revenue (R(p)) is p=25/4 or p=6.25
a maximum revenue R(p)=1875/4 or R(p)=468.75

 

Thank you pointing out my error. I appreciate every reply.