Find the absolute maximum and minimum values of f(x,y)=x^3+3y^2+4
 on the set D
 where D
 is the closed region bounded by y=0
 and y=5−5x^2. Along y=5−5x2
, f
 can be expressed as a function of one variable List all the points on this side of the boundary which could potentially be the absolute minimum or maximum on D
.

We are given the function \( f(x, y) = x^3 + 3y^2 + 4 \), and we are asked to find the absolute maximum and minimum values of this function on the closed region \( D \), which is bounded by the curves \( y = 0 \) and \( y = 5 - 5x^2 \).

### Steps to find the absolute maximum and minimum:

1. **Analyze the boundary of the region \( D \)**:
   - \( D \) is bounded by \( y = 0 \) (the x-axis) and \( y = 5 - 5x^2 \).
   - Along \( y = 5 - 5x^2 \), we can express \( f(x, y) \) as a function of \( x \), which allows us to reduce the problem to a single-variable function.
   
2. **Express \( f(x, y) \) on the curve \( y = 5 - 5x^2 \)**:
   Substitute \( y = 5 - 5x^2 \) into the function \( f(x, y) \):

   \[
   f(x, 5 - 5x^2) = x^3 + 3(5 - 5x^2)^2 + 4
   \]
   
   Simplifying this expression:
   \[
   f(x, 5 - 5x^2) = x^3 + 3(25 - 50x^2 + 25x^4) + 4
   \]
   \[
   f(x, 5 - 5x^2) = x^3 + 75 - 150x^2 + 75x^4 + 4
   \]
   \[
   f(x, 5 - 5x^2) = 75x^4 - 150x^2 + x^3 + 79
   \]

   Now we need to find the critical points of this function within the range \( x \in [-1, 1] \), because \( y = 5 - 5x^2 \) requires that \( x^2 \leq 1 \), meaning \( x \in [-1, 1] \).

3. **Find the derivative of \( f(x, 5 - 5x^2) \)** to locate the critical points:
   \[
   f'(x) = \frac{d}{dx}\left( 75x^4 - 150x^2 + x^3 + 79 \right)
   \]
   Applying the power rule:
   \[
   f'(x) = 300x^3 - 300x + 3x^2
   \]
   \[
   f'(x) = 3x(100x^2 - 100 + x)
   \]

4. **Solve for critical points** by setting \( f'(x) = 0 \):
   \[
   3x(100x^2 - 100 + x) = 0
   \]
   This equation has two factors:
   - \( x = 0 \)
   - \( 100x^2 - 100 + x = 0 \)
   
   Solve \( 100x^2 + x - 100 = 0 \) using the quadratic formula:
   \[
   x = \frac{-1 \pm \sqrt{1^2 - 4(100)(-100)}}{2(100)} = \frac{-1 \pm \sqrt{1 + 40000}}{200} = \frac{-1 \pm \sqrt{40001}}{200}
   \]
   The roots are approximately:
   \[
   x \approx \frac{-1 + 200.0025}{200} \approx 0.99999
   \quad \text{and} \quad x \approx \frac{-1 - 200.0025}{200} \approx -1.00001
   \]

5. **Check the critical points** and endpoints within the domain \( [-1, 1] \):
   - The critical points within the interval are \( x \approx 1 \) and \( x \approx -1 \), along with \( x = 0 \).
   - The endpoints of the domain are \( x = -1 \) and \( x = 1 \).

### Potential points on the boundary:
From the above analysis, the points on the boundary where the function might achieve its absolute minimum or maximum are:
- \( x = -1 \)
- \( x = 0 \)
- \( x = 1 \)

### Conclusion:
These are the points on the boundary that could potentially be the absolute minimum or maximum for the function on the region \( D \). Next, we would evaluate the function \( f(x, 5 - 5x^2) \) at these points to determine which gives the maximum and minimum values.

Would you like me to continue and compute the values at these points, or do you have any questions so far?

Find the absolute maximum and minimum values of f(x,y)=x^3+3y^2+4 on the set D where D is the closed region bounded by y=0 and y=5−5x^2. Along y=5−5x2, f can be expressed as a function of one variable. List all the points on this side of the boundary which could potentially be the absolute minimum or maximum on D.

Learn how to find the absolute maximum and minimum of the function f(x, y) = x^3 + 3y^2 + 4 on a closed region bounded by y = 0 and y = 5 - 5x^2. This guide covers steps for boundary analysis, critical point determination, and evaluating potential extreme values. Suitable for students in Grades 11-12.

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