Answer:
Marginal revenue at max revenue = 0
Instantaneous rate of change of revenue at 400 steam irons is 100
Step-by-step explanation:
We have the price-demand equation as
![x = 1000-2p](https://tex.z-dn.net/?f=x%20%3D%201000-2p)
Express p in terms of x
Adding 2p on both sides and subtracting x from both sides yields
![2p = 1000-x](https://tex.z-dn.net/?f=2p%20%3D%201000-x)
Dividing by 2 on both sides gives us
![p = 500-\frac{x}{2}](https://tex.z-dn.net/?f=p%20%3D%20500-%5Cfrac%7Bx%7D%7B2%7D)
The total revenue from selling x steam irons is given by
(1)
This revenue is maximized when the first differential is zero
First differential of
=
(2)
Setting this equal to zero and solving for
will give us the value of
at which the revenue is maximum and therefore the marginal revenue
Setting ![500-x = 0](https://tex.z-dn.net/?f=500-x%20%3D%200)
gives
x = 500
At this demand level marginal revenue is 0
ie the number of steam irons to be sold to maximize revenue
The revenue at this demand level is given by plugging in this value into equation (1)
![R_{max} = 500(500) - \frac{500^2}{2} = 500^2 - \frac{500^2}{2} = \frac{500^2}{2} = 125,000](https://tex.z-dn.net/?f=R_%7Bmax%7D%20%3D%20500%28500%29%20-%20%5Cfrac%7B500%5E2%7D%7B2%7D%20%3D%20500%5E2%20%20-%20%5Cfrac%7B500%5E2%7D%7B2%7D%20%3D%20%5Cfrac%7B500%5E2%7D%7B2%7D%20%3D%20125%2C000)
At x = 400, the instantaneous rate of change is given by plugging in this value into Equation (2)
![R'(400) = 500-400 = 100](https://tex.z-dn.net/?f=R%27%28400%29%20%3D%20500-400%20%3D%20100)
and the revenue at x = 400 is obtained by plugging in 400 for the value of x in Eq 1
![R(400) = (500)(400) - \frac{400^2}{2} = 200,000 - 80,000 = 120,000](https://tex.z-dn.net/?f=R%28400%29%20%3D%20%28500%29%28400%29%20-%20%5Cfrac%7B400%5E2%7D%7B2%7D%20%3D%20200%2C000%20-%2080%2C000%20%20%3D%20120%2C000)
Hope that helps