Find the maxima and minimum value of the function `y = 40/(3x^(4)+ 8x^(3)- 18x^(2) +60`.

Answer

To find the maximum and minimum values of the function

\[ y = \frac{40}{3x^4 + 8x^3 - 18x^2 + 60}, \]

we will first rewrite the function in terms of \( z \):

\[ z = \frac{1}{y} = \frac{3x^4 + 8x^3 - 18x^2 + 60}{40}. \]

### Step 1: Find the derivative of \( z \)

To find the critical points, we need to find the derivative \( \frac{dz}{dx} \):

\[ \frac{dz}{dx} = \frac{1}{40} \cdot (12x^3 + 24x^2 - 36x). \]

Setting the derivative equal to zero to find critical points:

\[ 12x^3 + 24x^2 - 36x = 0. \]

### Step 2: Factor the derivative

Factoring out the common term:

\[ 12x(x^2 + 2x - 3) = 0. \]

This gives us:

\[ 12x = 0 \quad \text{or} \quad x^2 + 2x - 3 = 0. \]

### Step 3: Solve for \( x \)

From \( 12x = 0 \), we have:

\[ x = 0. \]

For the quadratic equation \( x^2 + 2x - 3 = 0 \), we can factor it as:

\[ (x + 3)(x - 1) = 0. \]

Thus, the solutions are:

\[ x = -3 \quad \text{and} \quad x = 1. \]

### Step 4: Find the second derivative

Next, we need to find the second derivative \( \frac{d^2z}{dx^2} \) to determine the nature of these critical points:

\[ \frac{d^2z}{dx^2} = \frac{1}{40} \cdot (36x^2 + 48x - 36). \]

### Step 5: Evaluate the second derivative at critical points

1. **At \( x = 0 \)**:

\[ \frac{d^2z}{dx^2} = \frac{1}{40} \cdot (36(0)^2 + 48(0) - 36) = \frac{-36}{40} < 0 \quad \text{(Maximum)}. \]

2. **At \( x = 1 \)**:

\[ \frac{d^2z}{dx^2} = \frac{1}{40} \cdot (36(1)^2 + 48(1) - 36) = \frac{48}{40} > 0 \quad \text{(Minimum)}. \]

3. **At \( x = -3 \)**:

\[ \frac{d^2z}{dx^2} = \frac{1}{40} \cdot (36(-3)^2 + 48(-3) - 36) = \frac{36(9) - 144 - 36}{40} = \frac{324 - 180}{40} > 0 \quad \text{(Minimum)}. \]

### Step 6: Calculate the values of \( y \)

1. **At \( x = 0 \)**:

\[ y(0) = \frac{40}{3(0)^4 + 8(0)^3 - 18(0)^2 + 60} = \frac{40}{60} = \frac{2}{3}. \]

2. **At \( x = 1 \)**:

\[ y(1) = \frac{40}{3(1)^4 + 8(1)^3 - 18(1)^2 + 60} = \frac{40}{3 + 8 - 18 + 60} = \frac{40}{53}. \]

3. **At \( x = -3 \)**:

\[ y(-3) = \frac{40}{3(-3)^4 + 8(-3)^3 - 18(-3)^2 + 60} = \frac{40}{3(81) - 8(27) - 18(9) + 60} = \frac{40}{243 - 216 - 162 + 60} = \frac{40}{-75} = -\frac{8}{15}. \]

### Conclusion

- The maximum value of \( y \) occurs at \( x = 0 \) and is \( \frac{2}{3} \).
- The minimum value of \( y \) occurs at \( x = -3 \) and is \( -\frac{8}{15} \).