We are tasked with finding the maximum and minimum values of the function $f(x, y) = 4x + 4y$ on the ellipse given by the constraint $4x^2 + 25y^2 = 1$.

To solve this, we can apply the method of Lagrange multipliers. Here's the step-by-step process:

Step 1: Lagrange Multiplier Method

We introduce the Lagrange multiplier $\lambda$ and form the Lagrange function $\mathcal{L}(x, y, \lambda)$ as:

$\mathcal{L}(x, y, \lambda) = 4x + 4y - \lambda (4x^2 + 25y^2 - 1)$

Here, $4x^2 + 25y^2 - 1 = 0$ is the constraint equation (ellipse equation).

Step 2: Take Partial Derivatives

Next, we take the partial derivatives of $\mathcal{L}$ with respect to $x$, $y$, and $\lambda$, and set them equal to zero:

1. $\frac{\partial \mathcal{L}}{\partial x} = 4 - \lambda \cdot 8x = 0$
2. $\frac{\partial \mathcal{L}}{\partial y} = 4 - \lambda \cdot 50y = 0$
3. $\frac{\partial \mathcal{L}}{\partial \lambda} = -(4x^2 + 25y^2 - 1) = 0$

Step 3: Solve the System of Equations

From the first two equations:

$4 - 8\lambda x = 0 \quad \Rightarrow \quad \lambda = \frac{1}{2x}$ $4 - 50\lambda y = 0 \quad \Rightarrow \quad \lambda = \frac{1}{12.5y}$

Equating the two expressions for $\lambda$:

$\frac{1}{2x} = \frac{1}{12.5y}$

This simplifies to:

$12.5y = 2x \quad \Rightarrow \quad y = \frac{2x}{12.5} = \frac{x}{6.25}$

Step 4: Substitute into the Constraint Equation

Substitute $y = \frac{x}{6.25}$ into the constraint $4x^2 + 25y^2 = 1$:

$4x^2 + 25\left(\frac{x}{6.25}\right)^2 = 1$ $4x^2 + 25\left(\frac{x^2}{39.0625}\right) = 1$ $4x^2 + \frac{25x^2}{39.0625} = 1$

Multiply through by 39.0625 to eliminate the denominator:

$39.0625(4x^2) + 25x^2 = 39.0625$ $156.25x^2 + 25x^2 = 39.0625$ $181.25x^2 = 39.0625$ $x^2 = \frac{39.0625}{181.25} = \frac{1}{4.64}$ $x = \pm \frac{1}{\sqrt{4.64}} \approx \pm 0.465$

Step 5: Find the Corresponding $y$-Values

Since $y = \frac{x}{6.25}$, we substitute $x = 0.465$ and $x = -0.465$:

For $x = 0.465$, $y = \frac{0.465}{6.25} \approx 0.0744$.

For $x = -0.465$, $y = \frac{-0.465}{6.25} \approx -0.0744$.

Step 6: Evaluate $f(x, y)$

Now, evaluate $f(x, y) = 4x + 4y$ for both pairs $(x, y)$:

1. For $x = 0.465$, $y = 0.0744$: $f(0.465, 0.0744) = 4(0.465) + 4(0.0744) = 1.86 + 0.2976 = 2.1576$

2. For $x = -0.465$, $y = -0.0744$: $f(-0.465, -0.0744) = 4(-0.465) + 4(-0.0744) = -1.86 - 0.2976 = -2.1576$

Step 7: Conclusion

Thus, the maximum value of $f(x, y)$ on the ellipse is 2.1576 and the minimum value is -2.1576.

Would you like further details or have any questions?

Follow-up Questions:

1. How does the method of Lagrange multipliers help solve constrained optimization problems?
2. Can this method be applied to other types of constraints, such as non-linear or equality constraints?
3. How does the constraint shape affect the maximum and minimum values of the objective function?
4. What happens if we change the function $f(x, y)$ to something like $f(x, y) = x^2 + y^2$?
5. How would you approach this problem using parametric equations for the ellipse?

Tip:

For problems involving ellipses or other conic sections, it's often helpful to parameterize the constraint (e.g., using trigonometric functions) to simplify the optimization.