If `a^(2)+b^(2)+c^(2)=1` where, a,b,`cin R`, then the maximum value of `(4a-3b)^(2) + (5b-4c)^(2)+(3c-5a)^(2)` is

Options

A. 25
B. 50
C. 144
D. none of these

Explanation

To solve the problem of finding the maximum value of \( (4a - 3b)^2 + (5b - 4c)^2 + (3c - 5a)^2 \) given the constraint \( a^2 + b^2 + c^2 = 1 \), we can use the Cauchy-Schwarz inequality.

### Step-by-step Solution:

1. **Understand the Problem**: We need to maximize the expression \( (4a - 3b)^2 + (5b - 4c)^2 + (3c - 5a)^2 \) under the constraint \( a^2 + b^2 + c^2 = 1 \).

2. **Apply Cauchy-Schwarz Inequality**: According to the Cauchy-Schwarz inequality, for any vectors \( \mathbf{u} \) and \( \mathbf{v} \):
\[ |\mathbf{u} \cdot \mathbf{v}|^2 \leq (\mathbf{u} \cdot \mathbf{u})(\mathbf{v} \cdot \mathbf{v}) \]
We can consider the vectors:
\[ \mathbf{u} = (4a - 3b, 5b - 4c, 3c - 5a) \]
and
\[ \mathbf{v} = (1, 1, 1) \]

3. **Calculate the Dot Products**:
- First, compute \( \mathbf{u} \cdot \mathbf{u} \):
\[ (4a - 3b)^2 + (5b - 4c)^2 + (3c - 5a)^2 \]
- Next, compute \( \mathbf{v} \cdot \mathbf{v} \):
\[ 1^2 + 1^2 + 1^2 = 3 \]

4. **Set Up the Inequality**:
By applying Cauchy-Schwarz, we have:
\[ (4a - 3b + 5b - 4c + 3c - 5a)^2 \leq (1^2 + 1^2 + 1^2)((4a - 3b)^2 + (5b - 4c)^2 + (3c - 5a)^2) \]
This simplifies to:
\[ (0)^2 \leq 3((4a - 3b)^2 + (5b - 4c)^2 + (3c - 5a)^2) \]
which is trivially true.

5. **Maximize the Expression**:
We can also express the original expression as:
\[ (4a - 3b)^2 + (5b - 4c)^2 + (3c - 5a)^2 \leq 50(a^2 + b^2 + c^2) \]
Since \( a^2 + b^2 + c^2 = 1 \):
\[ (4a - 3b)^2 + (5b - 4c)^2 + (3c - 5a)^2 \leq 50 \]

6. **Conclusion**: The maximum value of \( (4a - 3b)^2 + (5b - 4c)^2 + (3c - 5a)^2 \) is \( 50 \).

### Final Answer:
The maximum value is \( \boxed{50} \).