C(x)=5000+2x, p(x)=10-0.001x, 0<x<10000, p(x) gives price

[helpmeout]

Hi everyone, Im new here and am having a problem with this one question:

C(x) = 5000 + 2x
p(x) = 10 - 0.001x 0<x<10000 p(x) describes the price.

Now I found the revenue function, but how do you find the marginal revenue, if x is not given? Do you plug in 10000?

 

[galactus]

Re: Newbie here: Need help on a question! Please Read!

Total revenue is \(\displaystyle xp(x)=x(10-\frac{x}{10000})=10x-\frac{x^{2}}{10000}\)

The marginal revenue is the extra revenue from selling one additional unit.

So, marginal revenue is the derivative of the total revenue.

If nothing is given, I would say differentiate the revenue function, set to 0 and solve for x.

This would give maximum marginal revenue.

Now, profit is P=R-C. \(\displaystyle P=(10x-\frac{x^{2}}{10000})-(5000+2x)=\frac{-x^{2}}{10000}+8x-5000\)

To find maximum marginal profit, differentiate, set to 0 and solve for x.

 

[Deleted member 4993]

Re: Newbie here: Need help on a question! Please Read!

Hi everyone, Im new here and am having a problem with this one question:

C(x) = 5000 + 2x
p(x) = 10 - 0.001x 0<x<10000 p(x) describes the price.

Now I found the revenue function, but how do you find the marginal revenue, if x is not given? Do you plug in 10000?

If do not know how to take derivative, note that the "Profit function" is a "parabola".

Find the vertex of the parabola for maximum value.

 

[galactus]

Re: Newbie here: Need help on a question! Please Read!

Good point, SK