Problem 14 - Calculus Early Transcendentals

Question

Lagrange multipliers in two variables Use Lagrange multipliers to find the maximum and minimum values of \(f\) (when they exist) subject to the given constraint. $$f(x, y)=x y+x+y \text { subject to } x^{2} y^{2}=4$$

Step-by-Step Solution

Answer

Answer: The maximum value is \(4 + 2\sqrt{2}\) achieved at \((\sqrt{2},\sqrt{2})\) and the minimum value is \(4 - 2\sqrt{2}\) achieved at \((-\sqrt{2},-\sqrt{2})\).

Step 1: 1. Set up the Lagrange equation

To apply the Lagrange multipliers method, we define a function \(L(x, y, \lambda) = f(x, y) - \lambda g(x, y)\), where \(g(x, y)\) is our constraint function. In this case, we have: $$L(x, y, \lambda) = (xy + x + y) - \lambda(x^2y^2 - 4)$$

Step 2: 2. Compute the gradients

Compute the gradient of \(L\) with respect to \(x\), \(y\), and \(\lambda\): $$\nabla L = \left(\frac{\partial L}{\partial x}, \frac{\partial L}{\partial y}, \frac{\partial L}{\partial \lambda}\right)$$ Calculating the partial derivatives, we get: $$\frac{\partial L}{\partial x} = y+1-2\lambda xy^2$$ $$\frac{\partial L}{\partial y} = x+1-2\lambda x^2y$$ $$\frac{\partial L}{\partial \lambda} = -(x^2y^2-4)$$

Step 3: 3. Set up the system of equations

To find the critical points, we set the gradient components to 0: $$\begin{cases} y+1-2\lambda xy^2 = 0 \\ x+1-2\lambda x^2y = 0 \\ -(x^2y^2-4) = 0 \end{cases}$$

Step 4: 4. Solve the system of equations

Solving the system of equations, we first see that the third equation simplifies to the constraint \(x^2y^2=4\). We express \(y\) from the first equation and \(x\) from the second equation: $$y = 2\lambda xy^2 - 1$$ $$x = 2\lambda x^2y - 1$$ Substitute these expressions into the constraint: $$(2\lambda x^2y - 1)^2(2\lambda xy^2 - 1)^2 = 4$$ We can rewrite the equation as: $$(2\lambda - 1)^2(x^2y)^2 = 1$$ Since we know from the constraint \(x^2y^2 = 4\), we can substitute it into the equation: $$(2\lambda - 1)^2 \cdot 4^2 = 1$$ And solve for \(\lambda\): $$\lambda = \frac{1}{4}, \frac{3}{4}$$ Using the expressions for \(x\) and \(y\), we find two pairs of critical points: $$(x_1, y_1) = \left(\sqrt{2},\sqrt{2}\right)$$ $$(x_2, y_2) = \left(-\sqrt{2},-\sqrt{2}\right)$$

Step 5: 5. Determine the maximum and minimum values

Now, we evaluate the function \(f(x, y)\) at the critical points: $$f(x_1, y_1) = (\sqrt{2})(\sqrt{2}) + \sqrt{2} + \sqrt{2} = 4 + 2\sqrt{2}$$ $$f(x_2, y_2) = (-\sqrt{2})(-\sqrt{2}) - \sqrt{2} - \sqrt{2} = 4 - 2\sqrt{2}$$ Comparing the function values, we find that the maximum value is \(4 + 2\sqrt{2}\), achieved at \((\sqrt{2},\sqrt{2})\), and the minimum value is \(4 - 2\sqrt{2}\), achieved at \((-\sqrt{2},-\sqrt{2})\).

Key Concepts

Constrained Optimization

Constrained optimization is a key concept in calculus, especially when it comes to finding the maximum or minimum values of a function while being restricted by a constraint. Imagine trying to find the highest point on a mountain while staying on a particular path. In mathematical terms, this path is the constraint, and the mountain height is your function. In the exercise given, our task is to maximize or minimize the function \( f(x, y) = xy + x + y \) subject to a constraint \( x^2y^2 = 4 \).
This type of problem is tackled using the method of Lagrange multipliers. Instead of examining the function alone, we introduce a new variable, \( \lambda \) (the Lagrange multiplier), which helps us combine the function with the constraint. This allows us to explore both the function and constraint simultaneously.
By focusing on the gradients (derivatives indicating the steepest direction of function increase), we find where they align, which leads us to potential maximum or minimum points that abide by the constraint. It’s like finding a balance between our desired goal and the limit imposed by the constraint.

Gradient Computation

The next crucial step in solving a constrained optimization problem is computing the gradient. The gradient of a function provides the direction of steepest ascent. By aligning the gradients of our target function and constraint, we identify points that might lead to a maximum or minimum.
For the exercise, the Lagrange function is given by \( L(x, y, \lambda) = xy + x + y - \lambda(x^2y^2 - 4) \). We need to calculate the gradient of this function, expressed as a vector consisting of partial derivatives with respect to each variable \(x,\) \(y,\) and \(\lambda.\)

• The partial derivative with respect to \(x\) is \( \frac{\partial L}{\partial x} = y + 1 - 2\lambda xy^2.\)
• The partial derivative with respect to \(y\) is \( \frac{\partial L}{\partial y} = x + 1 - 2\lambda x^2y.\)
• The partial derivative with respect to \(\lambda\) is \( \frac{\partial L}{\partial \lambda} = -(x^2y^2 - 4).\)

By setting these partial derivatives equal to zero, we derive a system of equations that will pinpoint our critical values. These are not just any points but those that potentially mark the highest or lowest point of the function within the constraints provided.

Partial Derivatives

Partial derivatives form the backbone of both optimization and gradient computation. When we compute a partial derivative, we see how a function changes as we alter one variable, while keeping others constant. It provides a component of the full picture of a function's change across dimensions.
Let’s break down how partial derivatives were employed in this exercise:

• We started by differentiating \(L\) in relation to \(x, y,\) and \(\lambda.\)
• The partial derivative with respect to \(x\) considers how the function changes solely when \(x\) changes. Here it was \(y+1-2\lambda xy^2,\) capturing the effect of changes in \(x.\)
• Similarly, the partial derivative with respect to \(y\) is \(x+1-2\lambda x^2y.\)
• Finally, differentiating with respect to \(\lambda\), \(- (x^2y^2 - 4),\) primarily addresses the constraint itself.

Setting these partial derivatives to zero highlights where the change in all directions equals zero, indicating potential optimal points (maximum or minimum). This concept is particularly useful in multivariable calculus, enabling us to assess how a function's value evolves within its domain.