Let m and n be positive integers and `x,y gt 0 and x+y =k,` where `k` is constant. Let `f (x,y) = x ^(m)y ^(n),` then: (a) `f (x,y) ` is maximum when `x= (mk)/(m+n)` (b) `f (x,y)` is maximuim where `x =y` (c) maximum value of `f (x,y)` is `(m^(n)n ^(m) k ^(m+n))/((m+n)^(m+n))` (d) maximum value of `f (x,y)` is `(k ^(m+n) m ^(m)n ^(n))/((m+n)^(m+n))`

Options

A. `f (x,y) ` is maximum when `x= (mk)/(m+n)`
B. `f (x,y)` is maximuim wheere `x =y`
C. maximum value of `f (x,y) is (m^(n)n ^(m) k ^(m+n))/((m+n)^(m+n))`
D. maximum vlaue of `f (x,y) is (k ^(m+n) m ^(m)n ^(n))/((m+n)^(m+n))`

Answer

To solve the problem, we need to find the maximum value of the function \( f(x, y) = x^m y^n \) given the constraint \( x + y = k \). We will use the method of substitution and differentiation to find the maximum value.

### Step 1: Substitute for \( y \)

Since we have the constraint \( x + y = k \), we can express \( y \) in terms of \( x \):

\[ y = k - x \]

### Step 2: Rewrite the function

Now, substitute \( y \) into the function \( f(x, y) \):

\[ f(x) = x^m (k - x)^n \]

### Step 3: Differentiate \( f(x) \)

To find the maximum, we need to differentiate \( f(x) \) with respect to \( x \) and set the derivative equal to zero:

\[ f'(x) = \frac{d}{dx} \left( x^m (k - x)^n \right) \]

Using the product rule, we have:

\[ f'(x) = m x^{m-1} (k - x)^n + x^m \cdot n (k - x)^{n-1} (-1) \]

This simplifies to:

\[ f'(x) = m x^{m-1} (k - x)^n - n x^m (k - x)^{n-1} \]

### Step 4: Set the derivative to zero

Setting \( f'(x) = 0 \):

\[ m x^{m-1} (k - x)^n = n x^m (k - x)^{n-1} \]

### Step 5: Factor out common terms

We can factor out \( x^{m-1} (k - x)^{n-1} \):

\[ x^{m-1} (k - x)^{n-1} \left( m (k - x) - n x \right) = 0 \]

Since \( x > 0 \) and \( k - x > 0 \), we can ignore the first factor, leading to:

\[ m (k - x) - n x = 0 \]

### Step 6: Solve for \( x \)

Rearranging gives:

\[ mk - mx = nx \]

Combining terms:

\[ mk = (m + n)x \]

Thus,

\[ x = \frac{mk}{m+n} \]

### Step 7: Find \( y \)

Using \( y = k - x \):

\[ y = k - \frac{mk}{m+n} = \frac{nk}{m+n} \]

### Step 8: Maximum value of \( f(x, y) \)

Now we substitute \( x \) and \( y \) back into the function to find the maximum value:

\[ f\left(\frac{mk}{m+n}, \frac{nk}{m+n}\right) = \left(\frac{mk}{m+n}\right)^m \left(\frac{nk}{m+n}\right)^n \]

This simplifies to:

\[ = \frac{m^m n^n k^{m+n}}{(m+n)^{m+n}} \]

### Conclusion

Thus, the maximum value of \( f(x, y) \) is:

\[ \frac{m^m n^n k^{m+n}}{(m+n)^{m+n}} \]

### Final Answers

- (a) \( f(x,y) \) is maximum when \( x = \frac{mk}{m+n} \) is **True**.
- (b) \( f(x,y) \) is maximum where \( x = y \) is **False**.
- (c) Maximum value of \( f(x,y) \) is \( \frac{(m^n n^m k^{m+n})}{(m+n)^{m+n}} \) is **False**.
- (d) Maximum value of \( f(x,y) \) is \( \frac{k^{m+n} m^m n^n}{(m+n)^{m+n}} \) is **True**.