If `x+y=1` then find the maximum value of the function `xy^2`

Answer

To find the maximum value of the function \( f(x, y) = xy^2 \) given the constraint \( x + y = 1 \), we can follow these steps:

### Step 1: Express \( y \) in terms of \( x \)
From the constraint \( x + y = 1 \), we can express \( y \) as:
\[ y = 1 - x \]

### Step 2: Substitute \( y \) in the function
Now, substitute \( y \) into the function \( f(x, y) \):
\[ f(x) = x(1 - x)^2 \]

### Step 3: Expand the function
Next, we expand the function:
\[ f(x) = x(1 - 2x + x^2) = x - 2x^2 + x^3 \]

### Step 4: Differentiate the function
To find the maximum value, we need to differentiate \( f(x) \) with respect to \( x \):
\[ f'(x) = 1 - 4x + 3x^2 \]

### Step 5: Set the derivative to zero
Set the derivative equal to zero to find the critical points:
\[ 3x^2 - 4x + 1 = 0 \]

### Step 6: Solve the quadratic equation
We can solve the quadratic equation using the quadratic formula:
\[ x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} = \frac{4 \pm \sqrt{(-4)^2 - 4 \cdot 3 \cdot 1}}{2 \cdot 3} = \frac{4 \pm \sqrt{16 - 12}}{6} = \frac{4 \pm 2}{6} \]
This gives us:
\[ x = 1 \quad \text{or} \quad x = \frac{1}{3} \]

### Step 7: Determine the nature of the critical points
To determine whether these points are maxima or minima, we can use the second derivative test. First, we find the second derivative:
\[ f''(x) = 6x - 4 \]

Now evaluate \( f''(x) \) at the critical points:
1. For \( x = 1 \):
\[ f''(1) = 6(1) - 4 = 2 \quad (\text{positive, so minimum}) \]
2. For \( x = \frac{1}{3} \):
\[ f''\left(\frac{1}{3}\right) = 6\left(\frac{1}{3}\right) - 4 = 2 - 4 = -2 \quad (\text{negative, so maximum}) \]

### Step 8: Find the corresponding \( y \) value
Now, we find the corresponding \( y \) value when \( x = \frac{1}{3} \):
\[ y = 1 - x = 1 - \frac{1}{3} = \frac{2}{3} \]

### Step 9: Calculate the maximum value of the function
Finally, substitute \( x \) and \( y \) back into the function to find the maximum value:
\[ f\left(\frac{1}{3}, \frac{2}{3}\right) = \frac{1}{3} \left(\frac{2}{3}\right)^2 = \frac{1}{3} \cdot \frac{4}{9} = \frac{4}{27} \]

Thus, the maximum value of the function \( xy^2 \) given the constraint \( x + y = 1 \) is:
\[ \boxed{\frac{4}{27}} \]