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To find the maximum and minimum values of the function \( f(x, y) = 5x^2 y \) subject to the constraint \( g(x, y) = x^2 + y^4 = 5 \), we can use the method of **Lagrange multipliers**. This approach involves setting up the following system of equations:

1. Define the Lagrange function:
   \[
   \mathcal{L}(x, y, \lambda) = f(x, y) - \lambda (g(x, y) - 5)
   \]
   where \( \lambda \) is the Lagrange multiplier.

2. Take the partial derivatives of \( \mathcal{L} \) with respect to \( x \), \( y \), and \( \lambda \), and set them equal to zero:
   \[
   \frac{\partial \mathcal{L}}{\partial x} = \frac{\partial f}{\partial x} - \lambda \frac{\partial g}{\partial x} = 0
   \]
   \[
   \frac{\partial \mathcal{L}}{\partial y} = \frac{\partial f}{\partial y} - \lambda \frac{\partial g}{\partial y} = 0
   \]
   \[
   \frac{\partial \mathcal{L}}{\partial \lambda} = g(x, y) - 5 = 0
   \]

Now, let's go through these steps in detail:

### Step 1: Compute Partial Derivatives
1. The partial derivatives of \( f(x, y) = 5x^2 y \) are:
   \[
   \frac{\partial f}{\partial x} = 10x y
   \]
   \[
   \frac{\partial f}{\partial y} = 5x^2
   \]

2. The partial derivatives of \( g(x, y) = x^2 + y^4 \) are:
   \[
   \frac{\partial g}{\partial x} = 2x
   \]
   \[
   \frac{\partial g}{\partial y} = 4y^3
   \]

### Step 2: Set Up Lagrange Equations
Using \( \frac{\partial f}{\partial x} = \lambda \frac{\partial g}{\partial x} \) and \( \frac{\partial f}{\partial y} = \lambda \frac{\partial g}{\partial y} \), we have:
1. \( 10x y = \lambda \cdot 2x \)
2. \( 5x^2 = \lambda \cdot 4y^3 \)
3. The constraint \( x^2 + y^4 = 5 \).

From the first equation, assuming \( x \neq 0 \), we can simplify to:
\[
\lambda = \frac{10y}{2} = 5y
\]

Using this in the second equation, we get:
\[
5x^2 = (5y) \cdot 4y^3
\]
\[
5x^2 = 20y^4
\]
\[
x^2 = 4y^4
\]

### Step 3: Substitute into the Constraint
Now substitute \( x^2 = 4y^4 \) into the constraint equation \( x^2 + y^4 = 5 \):
\[
4y^4 + y^4 = 5
\]
\[
5y^4 = 5
\]
\[
y^4 = 1
\]
Thus, \( y = \pm 1 \).

### Step 4: Solve for \( x \)
1. If \( y = 1 \):
   \[
   x^2 = 4 \cdot 1^4 = 4 \Rightarrow x = \pm 2
   \]

2. If \( y = -1 \):
   \[
   x^2 = 4 \cdot (-1)^4 = 4 \Rightarrow x = \pm 2
   \]

### Step 5: Calculate \( f(x, y) \) at Critical Points
Now we evaluate \( f(x, y) = 5x^2 y \) at the points \((2, 1)\), \((-2, 1)\), \((2, -1)\), and \((-2, -1)\):

1. \( f(2, 1) = 5 \cdot (2)^2 \cdot 1 = 20 \)
2. \( f(-2, 1) = 5 \cdot (-2)^2 \cdot 1 = 20 \)
3. \( f(2, -1) = 5 \cdot (2)^2 \cdot (-1) = -20 \)
4. \( f(-2, -1) = 5 \cdot (-2)^2 \cdot (-1) = -20 \)

### Conclusion
- The **maximum value** of \( f(x, y) \) is \( 20 \).
- The **minimum value** of \( f(x, y) \) is \( -20 \).

Would you like further details on this solution or have any questions?

### Related Questions
1. How does the method of Lagrange multipliers work for three-variable functions?
2. What is the geometric interpretation of Lagrange multipliers?
3. How do we identify saddle points in constrained optimization problems?
4. Can the Lagrange method be applied to inequality constraints?
5. What are alternative methods to find extreme values with constraints?

### Tip
When using Lagrange multipliers, always check if the constraint function is differentiable and if the constraint equation describes a smooth curve or surface.

Use Lagrange multipliers to find the maximum and minimum values of the function f(x, y) = 5x^2 y subject to the constraint x^2 + y^4 = 5.

Learn how to use Lagrange multipliers to determine the maximum and minimum values of the function f(x, y) = 5x^2 y with the constraint x^2 + y^4 = 5. This solution involves setting up and solving the Lagrange function, computing partial derivatives, and applying optimization techniques. Suitable for college-level calculus students.

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