If `a gt b gt0` then maximum value of
`(ab(a^2-b^2)sinxcosx)/(a^2sin^2x+b^2cos^2x),x in (0,pi//2)` is

Options

A. `a^2-b^2`
B. `(a^2-b^2)/2`
C. `(a^2+b^2)/2`
D. none of these

Answer

To find the maximum value of the expression

\[ \frac{ab(a^2 - b^2) \sin x \cos x}{a^2 \sin^2 x + b^2 \cos^2 x} \]

for \(x \in (0, \frac{\pi}{2})\) given that \(a > b > 0\), we can follow these steps:

### Step 1: Rewrite the Expression
We start by rewriting the expression to facilitate simplification. We can divide both the numerator and the denominator by \(\sin x \cos x\):

\[ \frac{ab(a^2 - b^2)}{\frac{a^2 \sin^2 x + b^2 \cos^2 x}{\sin x \cos x}} \]

This gives us:

\[ \frac{ab(a^2 - b^2)}{a^2 \tan x + b^2 \cot x} \]

### Step 2: Apply AM-GM Inequality
We can apply the Arithmetic Mean-Geometric Mean (AM-GM) inequality to the terms in the denominator:

\[ \frac{a^2 \tan x + b^2 \cot x}{2} \geq \sqrt{a^2 \tan x \cdot b^2 \cot x} \]

This simplifies to:

\[ \frac{a^2 \tan x + b^2 \cot x}{2} \geq \sqrt{a^2 b^2} = ab \]

### Step 3: Rearranging the Inequality
From the AM-GM inequality, we can rearrange it to find:

\[ a^2 \tan x + b^2 \cot x \geq 2ab \]

### Step 4: Substitute Back into the Expression
Now substituting this back into our expression, we have:

\[ \frac{ab(a^2 - b^2)}{a^2 \tan x + b^2 \cot x} \leq \frac{ab(a^2 - b^2)}{2ab} = \frac{a^2 - b^2}{2} \]

### Step 5: Conclusion
Thus, the maximum value of the given expression is:

\[ \frac{a^2 - b^2}{2} \]

### Final Answer
The maximum value of the expression is \(\frac{a^2 - b^2}{2}\).

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