1, just the fax, max inc has determined that the demand for its fax machines is Q=3000-1.5P.

Question

A, calculate the ponit price elasitcity of deman when P=$600.
B, at P=$600 what is the frim is marginal revenues ?
C, find the total revenues maximizing price and quantity for the frim ?

Answer 1

Let's start with the given demand function:

\[ Q = 3000 - 1.5P \]

where \( Q \) is the quantity demanded and \( P \) is the price.

### Part A: Calculate the point price elasticity of demand when \( P = 600 \)

The point price elasticity of demand is calculated using the formula:

\[ E_d = \frac{dQ}{dP} \times \frac{P}{Q} \]

First, calculate the derivative \( \frac{dQ}{dP} \):

\[ \frac{dQ}{dP} = -1.5 \]

Next, substitute \( P = 600 \) into the demand function to find \( Q \):

\[ Q = 3000 - 1.5 \times 600 = 3000 - 900 = 2100 \]

Now, plug in the values into the elasticity formula:

\[ E_d = -1.5 \times \frac{600}{2100} \]
\[ E_d = -1.5 \times \frac{2}{7} \]
\[ E_d = -\frac{3}{7} \approx -0.43 \]

So, the point price elasticity of demand when \( P = 600 \) is approximately \(-0.43\).

### Part B: Calculate the firm's marginal revenue when \( P = 600 \)

Marginal revenue (MR) is calculated by taking the derivative of the total revenue (TR) function. Total revenue is \( TR = P \times Q \).

First, express \( TR \) in terms of \( P \):

\[ Q = 3000 - 1.5P \]
\[ TR = P \times Q \]
\[ TR = P \times (3000 - 1.5P) \]
\[ TR = 3000P - 1.5P^2 \]

Next, take the derivative of \( TR \) with respect to \( P \):

\[ MR = \frac{d(TR)}{dP} = 3000 - 3P \]

Now, substitute \( P = 600 \):

\[ MR = 3000 - 3 \times 600 \]
\[ MR = 3000 - 1800 \]
\[ MR = 1200 \]

So, the firm's marginal revenue when \( P = 600 \) is \( \$1200 \).

### Part C: Find the total revenue-maximizing price and quantity for the firm

To maximize total revenue, we need to find the price at which marginal revenue is zero because total revenue is maximized when \( MR = 0 \).

Set \( MR \) to zero and solve for \( P \):

\[ 3000 - 3P = 0 \]
\[ 3P = 3000 \]
\[ P = 1000 \]

Now, find the corresponding quantity \( Q \) by substituting \( P = 1000 \) into the demand function:

\[ Q = 3000 - 1.5 \times 1000 \]
\[ Q = 3000 - 1500 \]
\[ Q = 1500 \]

So, the total revenue-maximizing price is \( \$1000 \) and the total revenue-maximizing quantity is \( 1500 \) units.