Find the coordinates of the maximum and minimum points of h(t) = 50 + 40t^3 - 5t^4 -4t^5

To find the maximum and minimum points of the function \( h(t) = 50 + 40t^3 - 5t^4 - 4t^5 \), we will first calculate its derivative, set the derivative equal to zero to find critical points, and then determine whether these points correspond to maxima or minima.

### Step 1: Find the derivative

The derivative \( h'(t) \) is calculated as follows:

\[
h'(t) = \frac{d}{dt}(50 + 40t^3 - 5t^4 - 4t^5) = 120t^2 - 20t^3 - 20t^4
\]

### Step 2: Set the derivative equal to zero

Set \( h'(t) = 0 \):

\[
120t^2 - 20t^3 - 20t^4 = 0
\]

Factor out \( 20t^2 \):

\[
20t^2(6 - t - t^2) = 0
\]

This gives us \( t = 0 \) or solving \( 6 - t - t^2 = 0 \).

### Step 3: Solve the quadratic equation

Rearranging the quadratic equation gives:

\[
t^2 + t - 6 = 0
\]

Using the quadratic formula \( t = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \):

\[
t = \frac{-1 \pm \sqrt{1^2 - 4(1)(-6)}}{2(1)} = \frac{-1 \pm \sqrt{1 + 24}}{2} = \frac{-1 \pm 5}{2}
\]

This yields two solutions:

\[
t = \frac{4}{2} = 2 \quad \text{and} \quad t = \frac{-6}{2} = -3
\]

### Step 4: Critical points

The critical points are \( t = 0, 2, -3 \).

### Step 5: Determine maxima and minima

To classify the critical points, we can use the second derivative test.

Calculate the second derivative \( h''(t) \):

\[
h''(t) = \frac{d}{dt}(120t^2 - 20t^3 - 20t^4) = 240t - 60t^2 - 80t^3
\]

Evaluate \( h''(t) \) at each critical point:

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

   \[
   h''(0) = 240(0) - 60(0^2) - 80(0^3) = 0 \quad \text{(inconclusive)}
   \]

2. **At \( t = 2 \)**:

   \[
   h''(2) = 240(2) - 60(2^2) - 80(2^3) = 480 - 240 - 640 = -400 \quad \text{(local maximum)}
   \]

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

   \[
   h''(-3) = 240(-3) - 60(-3^2) - 80(-3^3) = -720 - 540 + 2160 = 900 \quad \text{(local minimum)}
   \]

### Step 6: Calculate \( h(t) \) at critical points

Now we find the function values at these points:

1. \( h(0) = 50 + 0 = 50 \)
2. \( h(2) = 50 + 40(2^3) - 5(2^4) - 4(2^5) = 50 + 320 - 80 - 128 = 162 \)
3. \( h(-3) = 50 + 40(-3)^3 - 5(-3)^4 - 4(-3)^5 = 50 - 1080 - 405 + 972 = -463 \)

### Conclusion

The coordinates of the maximum and minimum points are:

- **Maximum Point**: \( (2, 162) \)
- **Minimum Point**: \( (-3, -463) \)

If you need further details or have any questions, feel free to ask! Here are some related questions to consider:

1. How do you apply the first derivative test?
2. What are the implications of the second derivative test?
3. How would you sketch the graph of \( h(t) \)?
4. How can you find inflection points for this function?
5. What is the significance of critical points in optimization problems?

**Tip**: Always check the endpoints of the domain if you're working within a specific interval.

Find the coordinates of the maximum and minimum points of h(t) = 50 + 40t^3 - 5t^4 - 4t^5.

Explore the process of finding the maximum and minimum points of the function h(t) = 50 + 40t^3 - 5t^4 - 4t^5. This detailed solution includes derivative calculations, critical points analysis, and results showing maximum at (2, 162) and minimum at (-3, -463). Suitable for Grades 11-12.

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