If `a^2+b^2+c^2+d^2 = 1`, then the maximum value of a.bc.d is :

Options

A. A)1
B. B)2
C. C)1/16
D. D)data insufficient

Answer

To find the maximum value of \( a \cdot b \cdot c \cdot d \) given that \( a^2 + b^2 + c^2 + d^2 = 1 \), we can use the method of equality in the Arithmetic Mean-Geometric Mean (AM-GM) inequality.

### Step-by-Step Solution:

1. **Understanding the Constraint**: We know that \( a^2 + b^2 + c^2 + d^2 = 1 \). This means that the values of \( a, b, c, \) and \( d \) are constrained such that their squares sum to 1.

2. **Using AM-GM Inequality**: The AM-GM inequality states that for non-negative numbers, the arithmetic mean is greater than or equal to the geometric mean. Thus, we can write:
\[ \frac{a^2 + b^2 + c^2 + d^2}{4} \geq \sqrt[4]{a^2 \cdot b^2 \cdot c^2 \cdot d^2} \]
Substituting the constraint into the inequality gives:
\[ \frac{1}{4} \geq \sqrt[4]{(abcd)^2} \]

3. **Simplifying the Inequality**: Raising both sides to the fourth power, we have:
\[ \left(\frac{1}{4}\right)^4 \geq (abcd)^2 \]
This simplifies to:
\[ \frac{1}{256} \geq (abcd)^2 \]

4. **Taking the Square Root**: Taking the square root of both sides, we get:
\[ \frac{1}{16} \geq abcd \]
This indicates that the maximum value of \( abcd \) is \( \frac{1}{16} \).

5. **Finding When Equality Holds**: The equality in the AM-GM inequality holds when all the numbers are equal. Therefore, we set:
\[ a^2 = b^2 = c^2 = d^2 \]
Let \( a = b = c = d = x \). Then:
\[ 4x^2 = 1 \implies x^2 = \frac{1}{4} \implies x = \frac{1}{2} \]

6. **Calculating the Maximum Value**: Substituting \( x \) back, we find:
\[ abcd = \left(\frac{1}{2}\right) \cdot \left(\frac{1}{2}\right) \cdot \left(\frac{1}{2}\right) \cdot \left(\frac{1}{2}\right) = \left(\frac{1}{2}\right)^4 = \frac{1}{16} \]

Thus, the maximum value of \( a \cdot b \cdot c \cdot d \) is \( \frac{1}{16} \).

### Conclusion:
The maximum value of \( a \cdot b \cdot c \cdot d \) given the constraint \( a^2 + b^2 + c^2 + d^2 = 1 \) is \( \frac{1}{16} \).