The base AB of a triangle is 1 and height h of C from AB is less than or equal to `1/2.` Then maximum vlaue of the 4 times the product of the altitude of the triangle is `"_______"`

Answer

To solve the problem, we need to find the maximum value of \(4\) times the product of the altitudes of a triangle with base \(AB = 1\) and height \(h\) from point \(C\) to base \(AB\) where \(h \leq \frac{1}{2}\).

### Step-by-Step Solution:

1. **Understanding the Area of the Triangle**:
The area \(A\) of triangle \(ABC\) can be expressed as:
\[ A = \frac{1}{2} \times \text{base} \times \text{height} = \frac{1}{2} \times 1 \times h = \frac{h}{2} \]
Since \(h \leq \frac{1}{2}\), the maximum area of the triangle is:
\[ A_{\text{max}} = \frac{1}{2} \times \frac{1}{2} = \frac{1}{4} \]

2. **Identifying the Altitudes**:
Let the other two sides of the triangle be \(x\) and \(y\). The altitudes corresponding to these sides can be expressed as:
\[ h_x = \frac{2A}{x} \quad \text{and} \quad h_y = \frac{2A}{y} \]

3. **Expressing the Product of Altitudes**:
The product of the altitudes is given by:
\[ h_x \cdot h_y = \left(\frac{2A}{x}\right) \cdot \left(\frac{2A}{y}\right) = \frac{4A^2}{xy} \]

4. **Substituting the Maximum Area**:
Substituting \(A_{\text{max}} = \frac{1}{4}\):
\[ h_x \cdot h_y = \frac{4 \left(\frac{1}{4}\right)^2}{xy} = \frac{4 \cdot \frac{1}{16}}{xy} = \frac{1}{4xy} \]

5. **Using the Triangle Inequality**:
By the triangle inequality, we know that:
\[ x + y > 1 \]
Let \(k = x + y - 1\), where \(k > 0\). Thus, we can rewrite \(x + y\) as:
\[ x + y = 1 + k \]

6. **Applying the AM-GM Inequality**:
By the AM-GM inequality:
\[ \frac{x + y}{2} \geq \sqrt{xy} \]
Substituting \(x + y = 1 + k\):
\[ \frac{1 + k}{2} \geq \sqrt{xy} \]
Squaring both sides gives:
\[ \left(\frac{1 + k}{2}\right)^2 \geq xy \]

7. **Finding the Maximum Value**:
We need to maximize \(4 \cdot h_x \cdot h_y\):
\[ 4 \cdot h_x \cdot h_y = 4 \cdot \frac{1}{4xy} = \frac{1}{xy} \]
From the earlier inequality, we have:
\[ xy \leq \left(\frac{1 + k}{2}\right)^2 \]
Therefore, we want to minimize \(xy\) to maximize \(\frac{1}{xy}\).

8. **Finding the Minimum of \(xy\)**:
The minimum value of \(xy\) occurs when \(x\) and \(y\) are equal, hence:
\[ x = y = \frac{1 + k}{2} \]
Thus, the minimum value of \(xy\) is:
\[ xy = \left(\frac{1 + k}{2}\right)^2 \]

9. **Substituting Back**:
If \(k = 0\), then \(x + y = 1\) and \(xy = \frac{1}{4}\). Thus:
\[ 4 \cdot h_x \cdot h_y \text{ max } = 4 \cdot \frac{1}{\frac{1}{4}} = 4 \]

### Conclusion:
Thus, the maximum value of \(4\) times the product of the altitudes of the triangle is:
\[ \boxed{2} \]