applications of derivative #3

[Cherry]

For the revenue function given by

R(x) = 2800x+8x^2-x^3

a. Find the maximum average revenue.
b. Show the (there is a line over the R) R(x) attains its manimum at an x - value where (there is a line over the R) R(x) = (there is a line over the MR) MR

 

[Guido]

Steps

Solution:

If I remember correctly, extreme points (max or min) happen where the derivative is zero. Derivative represents the slope of the tangent line, which would be zero (horizontal) at extreme points. So, you want to take the derivative (R bar(x)) and set it to zero. There should be two solutions. You can plug each into the original equation and one of them should be the max.

As for b, I don't know what MR represents. Can you make this clear for me?

 

[Guido]

Additional Notes....

To get an average, you need the range of x.

This is done by integrating the function between x = a and b and dividing this integral by b-a.

The function maximum is determined from the derivative

iff'n R(x) = 2800x+8x^2-x^3.

Then dR(x)/dx=2800+16x-3x^2

Set this equal to zero and solve for x:

2800+16x-3x^2=0

x=[16+/-sqrt(16^2+4*3*28000]/(2*3)=(16+/-184)/6

A negative x makes no sense. So

x=(16+184)/6=33.3333

Check to see iff'n this is a max and take the 2nd derivative

d^2R(x)/dx^2=16-6x and let x=33.3333

Yes, it's negative. So, it's a maximum.

So, the max revenue is at x=33.3333.

 

[pka]