If `x+y=20` and P=xy, then what is the maximum value of P?

Options

A. 100
B. 96
C. 84
D. 50

Answer

To find the maximum value of \( P = xy \) given the constraint \( x + y = 20 \), we can use the method of inequalities, specifically the Arithmetic Mean-Geometric Mean (AM-GM) inequality.

### Step-by-Step Solution:

1. **Set Up the Problem**:
We know that \( x + y = 20 \). We want to maximize \( P = xy \).

2. **Use the AM-GM Inequality**:
The Arithmetic Mean (AM) of two non-negative numbers \( x \) and \( y \) is given by:
\[ \text{AM} = \frac{x + y}{2} \]
The Geometric Mean (GM) is given by:
\[ \text{GM} = \sqrt{xy} \]
According to the AM-GM inequality, we have:
\[ \frac{x + y}{2} \geq \sqrt{xy} \]

3. **Substitute the Known Value**:
Substitute \( x + y = 20 \) into the AM-GM inequality:
\[ \frac{20}{2} \geq \sqrt{xy} \]
Simplifying this gives:
\[ 10 \geq \sqrt{xy} \]

4. **Square Both Sides**:
To eliminate the square root, square both sides:
\[ 10^2 \geq xy \]
This simplifies to:
\[ 100 \geq xy \]

5. **Conclusion**:
The maximum value of \( P = xy \) is \( 100 \). This maximum occurs when \( x = y \). Given \( x + y = 20 \), if \( x = y \), then:
\[ 2x = 20 \implies x = 10 \quad \text{and} \quad y = 10 \]
Therefore, \( P = 10 \times 10 = 100 \).

### Final Answer:
The maximum value of \( P = xy \) is \( 100 \).

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