If the demand function for a product is `p=(80-pi)/(4)`, where x is the number of units and p is the price per unit, the value of x for which the revenue will be maximum is

Options

A. 40
B. 20
C. 10
D. 80

Answer

To find the value of \( x \) for which the revenue will be maximum, we will follow these steps:

### Step 1: Write the Demand Function
The demand function is given as:
\[ p = \frac{80 - x}{4} \]
where \( p \) is the price per unit and \( x \) is the number of units.

### Step 2: Write the Revenue Function
Revenue \( R \) is defined as the product of price per unit and the number of units sold:
\[ R = p \cdot x \]
Substituting the demand function into the revenue function:
\[ R = \left(\frac{80 - x}{4}\right) \cdot x \]
This simplifies to:
\[ R = \frac{80x - x^2}{4} \]
or
\[ R = 20x - \frac{x^2}{4} \]

### Step 3: Differentiate the Revenue Function
To find the maximum revenue, we need to differentiate the revenue function with respect to \( x \):
\[ \frac{dR}{dx} = 20 - \frac{2x}{4} = 20 - \frac{x}{2} \]

### Step 4: Set the Derivative Equal to Zero
To find the critical points, set the derivative equal to zero:
\[ 20 - \frac{x}{2} = 0 \]
Solving for \( x \):
\[ \frac{x}{2} = 20 \implies x = 40 \]

### Step 5: Verify Maximum or Minimum
To determine whether this critical point is a maximum or minimum, we need to find the second derivative of the revenue function:
\[ \frac{d^2R}{dx^2} = -\frac{1}{2} \]
Since the second derivative is negative, this indicates that the revenue function has a maximum at \( x = 40 \).

### Final Answer
The value of \( x \) for which the revenue will be maximum is:
\[ \boxed{40} \]