Given that `x +y=2a`, where a is constant and that all values of x between 0 and 2a are equally likely, then show that the chance that `xy gt (3)/(4) a ^2,, is (1)/(2)`.

Answer

To solve the problem, we start with the given equation and the conditions provided.

### Step-by-Step Solution:

1. **Understanding the Problem**: We are given that \( x + y = 2a \), where \( a \) is a constant. We need to find the probability that \( xy > \frac{3}{4} a^2 \).

2. **Expressing \( y \) in terms of \( x \)**: From the equation \( x + y = 2a \), we can express \( y \) as:
\[ y = 2a - x \]

3. **Substituting \( y \) into the inequality**: We substitute \( y \) into the inequality \( xy > \frac{3}{4} a^2 \):
\[ x(2a - x) > \frac{3}{4} a^2 \]
This simplifies to:
\[ 2ax - x^2 > \frac{3}{4} a^2 \]

4. **Rearranging the inequality**: Rearranging gives:
\[ -x^2 + 2ax - \frac{3}{4} a^2 > 0 \]
Multiplying through by -1 (which reverses the inequality):
\[ x^2 - 2ax + \frac{3}{4} a^2 < 0 \]

5. **Finding the roots of the quadratic equation**: We can find the roots of the quadratic equation \( x^2 - 2ax + \frac{3}{4} a^2 = 0 \) using the quadratic formula:
\[ x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \]
Here, \( a = 1, b = -2a, c = \frac{3}{4} a^2 \):
\[ x = \frac{2a \pm \sqrt{(-2a)^2 - 4 \cdot 1 \cdot \frac{3}{4} a^2}}{2 \cdot 1} \]
Simplifying the discriminant:
\[ = \frac{2a \pm \sqrt{4a^2 - 3a^2}}{2} = \frac{2a \pm \sqrt{a^2}}{2} = \frac{2a \pm a}{2} \]
This gives us the roots:
\[ x = \frac{3a}{2} \quad \text{and} \quad x = \frac{a}{2} \]

6. **Finding the interval for \( x \)**: The quadratic \( x^2 - 2ax + \frac{3}{4} a^2 < 0 \) is negative between its roots:
\[ \frac{a}{2} < x < \frac{3a}{2} \]

7. **Calculating the probability**: The total range of \( x \) is from \( 0 \) to \( 2a \). The length of the interval where \( xy > \frac{3}{4} a^2 \) is:
\[ \frac{3a}{2} - \frac{a}{2} = a \]
The total length of the interval for \( x \) is:
\[ 2a - 0 = 2a \]
Thus, the probability is:
\[ P(xy > \frac{3}{4} a^2) = \frac{a}{2a} = \frac{1}{2} \]

### Conclusion:
We have shown that the probability that \( xy > \frac{3}{4} a^2 \) is \( \frac{1}{2} \).