Problem 13 - 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. $$f(x, y)=y^{2}-4 x^{2} \text { subject to } x^{2}+2 y^{2}=4$$

Step-by-Step Solution

Answer

To summarize, the maximum and minimum values of the function \(f(x, y) = y^2 - 4x^2\) subject to the constraint \(x^2 + 2y^2 = 4\) are found to be 1 and -16, respectively, using the method of Lagrange multipliers. The maximum occurs at the critical points (0, 1) and (0, -1), while the minimum occurs at the critical points (2, 0) and (-2, 0).

Step 1: Write down the function and constraint

Write the function f(x, y) and the constraint g(x, y) that we want to optimize: $$f(x, y) = y^2 - 4x^2$$ $$g(x, y) = x^2 + 2y^2 = 4$$

Step 2: Compute the gradient of f and g

Finding the gradient (the vector of first partial derivatives) of f and the gradient of g: $$\nabla f(x, y) = \frac{\partial f}{\partial x}\hat{\imath} + \frac{\partial f}{\partial y}\hat{\jmath} = (-8x)\hat{\imath} + (2y)\hat{\jmath}$$ $$\nabla g(x, y) = \frac{\partial g}{\partial x}\hat{\imath} + \frac{\partial g}{\partial y}\hat{\jmath} = (2x)\hat{\imath} + (4y)\hat{\jmath}$$

Step 3: Apply method of Lagrange multipliers

The method of Lagrange multipliers states that at the optimal points (the max or min subject to the constraint), the gradients of f and g are parallel, meaning: $$\nabla f(x, y) = \lambda \nabla g(x, y)$$ With \(\lambda\) being the Lagrange multiplier. Setting the components equal, we get the following system of equations: $$-8x = 2x\lambda$$ $$(2y) = 4y\lambda$$ $$x^2 + 2y^2 = 4$$

Step 4: Solve the system of equations

Solve the system of equations to find the optimal points (x, y) and their corresponding values of \(\lambda\): From the second equation, we can derive that either (y=0) or (\(\lambda=\frac{1}{2}\)). If y = 0, from the constraint, we have x = ±2, which are two critical points. If \(\lambda = \frac{1}{2}\), substitute this value into the first equation to get x = 0, which gives us y = ±1. So we have found four critical points: (±2, 0), and (0, ±1).

Step 5: Evaluate f at critical points

Evaluate the function f at the critical points to find the maximum and minimum values: $$f(2, 0) = -16$$ $$f(-2, 0) = -16$$ $$f(0, 1) = 1$$ $$f(0, -1) = 1$$ The maximum and minimum values of f subject to the constraint are 1 and -16, respectively.

Key Concepts

Multivariable Calculus

Multivariable calculus is a branch of mathematics that deals with functions involving more than one variable. It extends the concepts of differentiation and integration from one-dimensional space to higher dimensions. In this context, you might be dealing with functions like \(f(x, y)\) or \(f(x, y, z)\), where variables influence the outcome of the function.

Key components of multivariable calculus include partial derivatives and gradients. Partial derivatives help us understand how a function changes as one of the variables is varied, keeping the others constant. The gradient is a vector that points in the direction of the steepest ascent of the function. It is composed of the partial derivatives with respect to each variable. Recognizing these components is crucial when exploring problems involving multiple variables.

Optimization Problem

An optimization problem aims to find the best solution from a set of feasible solutions. In the context of calculus, it pertains to finding the extreme values (maximum or minimum) of a function.

Here, the goal is to maximize or minimize a function subject to certain conditions or constraints. This can be achieved through comparing function values at critical points found by setting derivatives equal to zero or by using other methods like Lagrange multipliers.

Optimization problems are critical in a wide array of fields such as economics, engineering, and operations research, where decisions must be optimized under given conditions.

Constrained Optimization

Constrained optimization refers to optimizing a function while respecting one or several restrictions or constraints. These constraints are often expressed in the form of equations or inequalities.

In our exercise, the function \(f(x, y) = y^2 - 4x^2\) is optimized subject to the constraint \(x^2 + 2y^2 = 4\). This constraint describes an ellipse in the \(xy\)-plane, and the goal is to find where on this ellipse the function reaches its highest or lowest values.

Constrained optimization problems often use techniques like the method of Lagrange multipliers, which allows us to incorporate constraints directly into the optimization process.

Gradient Vectors

Gradient vectors are a crucial concept in multivariable calculus, especially in optimization. A gradient vector represents the direction and rate of fastest increase of a scalar function. For a function \(f(x, y)\), the gradient \(abla f(x, y)\) is a vector composed of its partial derivatives: \(abla f(x, y) = \left( \frac{\partial f}{\partial x}, \frac{\partial f}{\partial y} \right)\).

In the method of Lagrange multipliers, gradient vectors play a vital role. The method relies on the principle that at an optimal point constrained by an equation \(g(x, y) = 0\), the gradient of \(f\) and the gradient of the constraint \(g\) are parallel. This means \(abla f(x, y) = \lambda abla g(x, y)\), where \(\lambda\) is a scalar known as the Lagrange multiplier. This equation helps find critical points that might be maximum or minimum values for the original function under the given constraint.