If z is a complex number lying in the first quadrant such that `"Re"(z)+"Im"(z)=3`, then the maximum value of `"{Re"(z)}^(2)"Im"(z)`, is

Options

A. 1
B. 2
C. 3
D. 4

Explanation

To solve the problem, we need to find the maximum value of the expression \( P = (\text{Re}(z))^2 \cdot \text{Im}(z) \) given that \( \text{Re}(z) + \text{Im}(z) = 3 \) and that \( z \) lies in the first quadrant.

Let's denote:
- \( \text{Re}(z) = x \)
- \( \text{Im}(z) = y \)

From the condition given, we have:
\[ x + y = 3 \]
This implies:
\[ y = 3 - x \]

Now, we can substitute \( y \) into the expression for \( P \):
\[ P = x^2 \cdot y = x^2 \cdot (3 - x) \]
This simplifies to:
\[ P = 3x^2 - x^3 \]

Next, we need to find the maximum value of \( P \). To do this, we will differentiate \( P \) with respect to \( x \) and set the derivative equal to zero:
\[ \frac{dP}{dx} = \frac{d}{dx}(3x^2 - x^3) = 6x - 3x^2 \]

Setting the derivative equal to zero:
\[ 6x - 3x^2 = 0 \]
Factoring out \( 3x \):
\[ 3x(2 - x) = 0 \]
This gives us two critical points:
\[ x = 0 \quad \text{or} \quad x = 2 \]

Next, we need to determine which of these points gives a maximum value. We will check the second derivative:
\[ \frac{d^2P}{dx^2} = 6 - 6x \]

Now we evaluate the second derivative at our critical points:
1. For \( x = 0 \):
\[ \frac{d^2P}{dx^2} = 6 - 6(0) = 6 \quad (\text{positive, indicating a minimum}) \]
2. For \( x = 2 \):
\[ \frac{d^2P}{dx^2} = 6 - 6(2) = 6 - 12 = -6 \quad (\text{negative, indicating a maximum}) \]

Since \( x = 2 \) gives a maximum, we now find the value of \( P \) at this point:
\[ y = 3 - x = 3 - 2 = 1 \]
Thus,
\[ P = (2)^2 \cdot (1) = 4 \]

Therefore, the maximum value of \( P \) is:
\[ \boxed{4} \]