If sum of two numbers is 3, then maximum value of the product of first and the square of second is

Options

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

Explanation

To solve the problem, we need to find the maximum value of the product of the first number \( x \) and the square of the second number \( y \), given that their sum is 3. Let's go through the solution step by step.

### Step 1: Define the Variables
Let the two numbers be:
- First number: \( x \)
- Second number: \( y \)

According to the problem, we have the equation:
\[ x + y = 3 \]

### Step 2: Express the Product
We want to maximize the product \( P \) defined as:
\[ P = x \cdot y^2 \]

### Step 3: Substitute for \( x \)
From the equation \( x + y = 3 \), we can express \( x \) in terms of \( y \):
\[ x = 3 - y \]

### Step 4: Substitute \( x \) in the Product
Now substitute \( x \) into the product \( P \):
\[ P = (3 - y) \cdot y^2 \]
Expanding this gives:
\[ P = 3y^2 - y^3 \]

### Step 5: Differentiate to Find Critical Points
To find the maximum value of \( P \), we need to differentiate it with respect to \( y \):
\[ \frac{dP}{dy} = 6y - 3y^2 \]

### Step 6: Set the Derivative to Zero
Now, set the derivative equal to zero to find the critical points:
\[ 6y - 3y^2 = 0 \]
Factoring out \( 3y \):
\[ 3y(2 - y) = 0 \]
This gives us two critical points:
\[ y = 0 \quad \text{or} \quad y = 2 \]

### Step 7: Determine the Nature of Critical Points
We need to determine whether these critical points correspond to a maximum or minimum. We do this by finding the second derivative:
\[ \frac{d^2P}{dy^2} = 6 - 6y \]

Evaluate the second derivative at the critical points:
1. For \( y = 0 \):
\[ \frac{d^2P}{dy^2} = 6 - 6(0) = 6 \quad (\text{positive, hence minimum}) \]

2. For \( y = 2 \):
\[ \frac{d^2P}{dy^2} = 6 - 6(2) = -6 \quad (\text{negative, hence maximum}) \]

### Step 8: Find Corresponding \( x \) Value
Now, we need to find the corresponding value of \( x \) when \( y = 2 \):
\[ x = 3 - y = 3 - 2 = 1 \]

### Step 9: Calculate Maximum Product
Now, substitute \( x \) and \( y \) back into the product \( P \):
\[ P = x \cdot y^2 = 1 \cdot (2^2) = 1 \cdot 4 = 4 \]

### Conclusion
Thus, the maximum value of the product \( x \cdot y^2 \) is:
\[ \boxed{4} \]