Maximize average revenue

[Ian McPherson]

The function R(with a line over it)(x) / x defines the average revenue for selling x units. For
R(x) = 2000x + 20x^2 - x^3

a) find the maximum average revenue
b)show that R(with a line over it) (x) attains its maxumum at an x-value where R(line over it)(x) = MR (Both with a line over it
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I'm not exactly sure where to begin, but I gave maximization my best shot.
*I found the derivative of R(x) to be -3x^2 +40x + 2000
However, the problem does not factor into workable numbers, nor does setting it = to zero to find critical numbers yield anything for me that looks workable. I set R'(x) = 0 and got 3x^2 = 40x + 2000, and divided both sides giving me (3/40)x^2 = x + 2000, but i don't believe that I'm taking the correct approach.

*I also tried putting R(x) over x as the avereage revenue funtion given looks.
(-x^3 + 20x^2 +2000x)/x, move the denominator to the top giving -x^3 + 20x^2 +2000x + x^-1. The derivative of that is -3x^2 +40x + 2000 -1x^-2. Move the -1x^-2 back down to get (-3x^2+40x+2000) / -x^2. Again, not sure what to do

I know it's confusing what i wrote, but are either of these two attempts the correct way to work the problem? I'm very lost, so if either are (doubtfully) correct, could you please explain what to do next? Thanks,
Ian

 

[srmichael]

First you have to find the average revenue function which is defined as \(\displaystyle \frac{R(x)}{x}\). Therefore:

\(\displaystyle \frac{R(x)}{x}=\frac{2000x+20x^2-x^3}{x}=2000+20x-x^2\)

Take the derivative of this and get: \(\displaystyle 20-2x\)

Setting this equal to 0 gives you x=10 units.

Plugging x=10 back into \(\displaystyle 2000+20x-x^2\) yields an average revenue of $2,100

 

[Ian McPherson]

I never thought to divide the denominator into the numerator before taking the derivative. I kept trying to move the denominator up and give it a negative exponent, making things very complicated. Thank you.:-D

 

[srmichael]

I never thought to divide the denominator into the numerator before taking the derivative. I kept trying to move the denominator up and give it a negative exponent, making things very complicated. Thank you.:-D

You're welcome!