Question

The price p (in dollars) and the demand x for a particular steam iron are related by the equation x=1000-2p.

Answer 1

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  

$x = 1000-2p$

Express p in terms of x

Adding 2p on both sides and subtracting x from both sides yields

$2p = 1000-x$

Dividing by 2 on both sides gives us

$p = 500-\frac{x}{2}$

The total revenue from selling x steam irons is given by

$R(x) = px = (500-\frac{x}{2})x = 500x - \frac{x^2}{2}$    (1)

This revenue is maximized  when the first differential is zero

First differential of $R(x)$ = $R'(x) = \frac{d}{dx} (500x - \frac{x^2}{2}) = 500 - \frac{2x}{2} = 500 - x$  (2)

Setting this equal to zero and solving for $x$ will give us the value of $x$ at which the revenue is maximum and therefore the marginal revenue

Setting $500-x = 0$

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)

$R_{max} = 500(500) - \frac{500^2}{2} = 500^2 - \frac{500^2}{2} = \frac{500^2}{2} = 125,000$

At x = 400, the instantaneous rate of change is given by plugging in this value into Equation (2)

$R'(400) = 500-400 = 100$

and the revenue at x = 400 is obtained by plugging in 400 for the value of x in Eq 1

$R(400) = (500)(400) - \frac{400^2}{2} = 200,000 - 80,000 = 120,000$

Hope that helps