Find the points of absolute maximum and minimum of each of the following :
`y=x(1+10x-x^(2)),3 le x le 9`

Answer

To find the points of absolute maximum and minimum of the function \( y = x(1 + 10x - x^2) \) in the interval \( [3, 9] \), we will follow these steps:

### Step 1: Differentiate the function
First, we need to find the derivative of the function \( y \) to locate the critical points.

\[ y = x(1 + 10x - x^2) \]

Using the product rule, we differentiate:

\[ \frac{dy}{dx} = (1 + 10x - x^2) + x(10 - 2x) \]

Simplifying this gives:

\[ \frac{dy}{dx} = 1 + 10x - x^2 + 10x - 2x^2 \]
\[ = 1 + 20x - 3x^2 \]

### Step 2: Set the derivative equal to zero
To find the critical points, we set the derivative equal to zero:

\[ 1 + 20x - 3x^2 = 0 \]

Rearranging gives:

\[ 3x^2 - 20x - 1 = 0 \]

### Step 3: Solve the quadratic equation
We can use the quadratic formula \( x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \) where \( a = 3, b = -20, c = -1 \):

\[ x = \frac{-(-20) \pm \sqrt{(-20)^2 - 4 \cdot 3 \cdot (-1)}}{2 \cdot 3} \]
\[ = \frac{20 \pm \sqrt{400 + 12}}{6} \]
\[ = \frac{20 \pm \sqrt{412}}{6} \]

Calculating \( \sqrt{412} \) gives approximately \( 20.29 \):

\[ x = \frac{20 \pm 20.29}{6} \]

Calculating the two potential critical points:

1. \( x = \frac{40.29}{6} \approx 6.715 \)
2. \( x = \frac{-0.29}{6} \approx -0.048 \) (not in the interval)

### Step 4: Evaluate the function at the endpoints and critical points
Now we evaluate \( y \) at the endpoints \( x = 3 \) and \( x = 9 \), and at the critical point \( x \approx 6.715 \).

1. **At \( x = 3 \)**:
\[ y(3) = 3(1 + 10 \cdot 3 - 3^2) = 3(1 + 30 - 9) = 3(22) = 66 \]

2. **At \( x = 9 \)**:
\[ y(9) = 9(1 + 10 \cdot 9 - 9^2) = 9(1 + 90 - 81) = 9(10) = 90 \]

3. **At \( x \approx 6.715 \)**:
\[ y(6.715) = 6.715(1 + 10 \cdot 6.715 - (6.715)^2) \]
First, calculate \( 10 \cdot 6.715 \approx 67.15 \) and \( (6.715)^2 \approx 45.09 \):
\[ y(6.715) \approx 6.715(1 + 67.15 - 45.09) = 6.715(23.06) \approx 154.84 \]

### Step 5: Compare the values
Now we compare the values obtained:

- \( y(3) = 66 \)
- \( y(9) = 90 \)
- \( y(6.715) \approx 154.84 \)

### Conclusion
The absolute maximum value occurs at \( x \approx 6.715 \) with \( y \approx 154.84 \), and the absolute minimum value occurs at \( x = 3 \) with \( y = 66 \).

### Final Answer
- Absolute Maximum: \( (6.715, 154.84) \)
- Absolute Minimum: \( (3, 66) \)