The price P for the demand D is given as `P = 183 + 120D − 3D^(2)` then the value of D for which price is increasing, is:

Options

A. `D lt 60`
B. `D gt 60`
C. `D lt 20`
D. `D gt 20`

Answer

To find the value of demand \( D \) for which the price \( P \) is increasing, we start with the given equation:

\[ P = 183 + 120D - 3D^2 \]

### Step 1: Differentiate P with respect to D

To determine when the price is increasing, we need to find the derivative of \( P \) with respect to \( D \) and set it greater than zero:

\[ \frac{dP}{dD} = \frac{d}{dD}(183 + 120D - 3D^2) \]

### Step 2: Calculate the derivative

Differentiating each term:

- The derivative of a constant (183) is 0.
- The derivative of \( 120D \) is \( 120 \).
- The derivative of \( -3D^2 \) is \( -6D \).

So we have:

\[ \frac{dP}{dD} = 0 + 120 - 6D = 120 - 6D \]

### Step 3: Set the derivative greater than zero

To find the values of \( D \) for which the price is increasing, we set the derivative greater than zero:

\[ 120 - 6D > 0 \]

### Step 4: Solve the inequality

Rearranging the inequality:

\[ 120 > 6D \]

Dividing both sides by 6:

\[ 20 > D \]

This can be rewritten as:

\[ D < 20 \]

### Conclusion

The value of \( D \) for which the price is increasing is:

\[ D < 20 \]