If `x+y=28`, then the maximum value of `x^(3)y^(4)` is

Options

A. `1234`
B. `4321`
C. `4^(3)24^(4)`
D. `12^(3)16^(4)`

Answer

To find the maximum value of \( x^3 y^4 \) given that \( x + y = 28 \), we can follow these steps:

### Step 1: Express \( y \) in terms of \( x \)
Since we know that \( x + y = 28 \), we can express \( y \) as:
\[ y = 28 - x \]

### Step 2: Substitute \( y \) into the expression
Now, we substitute \( y \) into the expression \( x^3 y^4 \):
\[ x^3 y^4 = x^3 (28 - x)^4 \]

### Step 3: Differentiate the expression
Next, we need to differentiate the expression \( x^3 (28 - x)^4 \) with respect to \( x \) and set the derivative equal to zero to find the critical points:
\[ \frac{d}{dx}(x^3 (28 - x)^4) = 0 \]

Using the product rule:
\[ \frac{d}{dx}(x^3) \cdot (28 - x)^4 + x^3 \cdot \frac{d}{dx}((28 - x)^4) = 0 \]

Calculating the derivatives:
\[ 3x^2 (28 - x)^4 + x^3 \cdot 4(28 - x)^3 \cdot (-1) = 0 \]

This simplifies to:
\[ 3x^2 (28 - x)^4 - 4x^3 (28 - x)^3 = 0 \]

### Step 4: Factor the equation
We can factor out common terms:
\[ (28 - x)^3 \left( 3(28 - x) - 4x \right) = 0 \]

This gives us:
\[ (28 - x)^3 = 0 \quad \text{or} \quad 3(28 - x) - 4x = 0 \]

From \( (28 - x)^3 = 0 \), we find \( x = 28 \) (not valid since \( y \) would be 0).

From \( 3(28 - x) - 4x = 0 \):
\[ 84 - 3x - 4x = 0 \implies 84 = 7x \implies x = 12 \]

### Step 5: Find \( y \)
Now, substituting \( x = 12 \) back into the equation for \( y \):
\[ y = 28 - x = 28 - 12 = 16 \]

### Step 6: Calculate the maximum value
Now we can calculate the maximum value of \( x^3 y^4 \):
\[ x^3 y^4 = 12^3 \cdot 16^4 \]

Calculating:
\[ 12^3 = 1728 \quad \text{and} \quad 16^4 = 65536 \]
Thus,
\[ x^3 y^4 = 1728 \cdot 65536 \]

### Final Result
Calculating \( 1728 \cdot 65536 \) gives us the maximum value.