If are positive integers, maximum value of `x^(m) (a-x)^(n)" in (0,a) is :"`

Options

A. `m^(m) (a-m)^(n)`
B. `m^(m) n^(n)`
C. `(m^(m) n^(n) a^(m+n))/((m+n)^(m+n))`
D. None of these

Answer

To find the maximum value of the function \( f(x) = x^m (a - x)^n \) for positive integers \( m \) and \( n \) in the interval \( (0, a) \), we will follow these steps:

### Step 1: Define the function
We start with the function:
\[ f(x) = x^m (a - x)^n \]

### Step 2: Differentiate the function
To find the maximum value, we need to differentiate \( f(x) \) with respect to \( x \):
\[ f'(x) = \frac{d}{dx}(x^m (a - x)^n) \]

Using the product rule:
\[ f'(x) = m x^{m-1} (a - x)^n + x^m \cdot n (a - x)^{n-1} \cdot (-1) \]
This simplifies to:
\[ f'(x) = m x^{m-1} (a - x)^n - n x^m (a - x)^{n-1} \]

### Step 3: Set the derivative to zero
To find critical points, set \( f'(x) = 0 \):
\[ m x^{m-1} (a - x)^n = n x^m (a - x)^{n-1} \]

### Step 4: Rearranging the equation
Rearranging gives:
\[ m (a - x) = n x \]
This can be rearranged to:
\[ ma - mx = nx \]
\[ ma = (m + n)x \]
Thus, we find:
\[ x = \frac{ma}{m+n} \]

### Step 5: Substitute back to find maximum value
Now substitute \( x = \frac{ma}{m+n} \) back into the function \( f(x) \):
\[ f\left(\frac{ma}{m+n}\right) = \left(\frac{ma}{m+n}\right)^m \left(a - \frac{ma}{m+n}\right)^n \]

Calculating \( a - \frac{ma}{m+n} \):
\[ a - \frac{ma}{m+n} = \frac{(m+n)a - ma}{m+n} = \frac{na}{m+n} \]

Thus, we have:
\[ f\left(\frac{ma}{m+n}\right) = \left(\frac{ma}{m+n}\right)^m \left(\frac{na}{m+n}\right)^n \]

### Step 6: Simplify the expression
Now, simplifying:
\[ f\left(\frac{ma}{m+n}\right) = \frac{m^m a^m}{(m+n)^m} \cdot \frac{n^n a^n}{(m+n)^n} \]
\[ = \frac{m^m n^n a^{m+n}}{(m+n)^{m+n}} \]

### Conclusion
The maximum value of \( f(x) \) in the interval \( (0, a) \) is:
\[ \frac{m^m n^n a^{m+n}}{(m+n)^{m+n}} \]