Problem 47 - Calculus Early Transcendentals

Question

Find the absolute maximum and minimum values of the following functions on the given set \(R\). $$\begin{aligned} &f(x, y)=2 x^{2}-4 x+3 y^{2}+2\\\ &R=\left\\{(x, y):(x-1)^{2}+y^{2} \leq 1\right\\} \end{aligned}$$

Step-by-Step Solution

Answer

**Question:** Find the absolute maximum and minimum values of the function \(f(x, y)=2x^2 - 4x + 3y^2 + 2\) on the set \(R=\{(x, y): (x-1)^2 + y^2 \leq 1\}\).

Step 1: Compute the gradient of \(f(x, y)\)

First, we need to compute the partial derivatives of \(f(x, y)\) with respect to \(x\) and \(y\). That is $$\begin{aligned} \frac{\partial f}{\partial x} = 4x - 4\\ \frac{\partial f}{\partial y} = 6y \end{aligned}$$

Step 2: Set the gradient equal to zero vector

Now, set the gradient equal to a zero vector to find the critical points: $$\begin{aligned} 4x - 4 &= 0 \\ 6y &= 0 \end{aligned}$$

Step 3: Solve for critical points

Solve the system of equations for \(x\) and \(y\) to find the critical points: $$\begin{aligned} x &= 1 \\ y &= 0 \end{aligned}$$ So there is one critical point \((1, 0)\).

Step 4: Check values of the function at the critical point and boundary points

Now, check the value of the function at the critical point \((1, 0)\) and at the boundary points, which lie on the circle \((x-1)^2 + y^2 = 1\). At the critical point \((1, 0)\), we have: $$f(1, 0) = 2$$ At the boundary points, let \(x-1 = \cos \theta\) and \(y = \sin \theta\) for \(0 \leq \theta \leq 2\pi\). Thus, $$\begin{aligned} f(\cos \theta +1, \sin \theta) &= 2(\cos \theta + 1)^2 -4(\cos \theta + 1) + 3(\sin \theta)^2 + 2\\ \end{aligned}$$

Step 5: Find the maximum and minimum values of \(f(x, y)\) on the boundary

Differentiate the expression of the function at the boundary point with respect to \(\theta\) and set it equal to zero: $$\frac{d}{d\theta} \left[2(\cos \theta + 1)^2 -4(\cos \theta + 1) + 3(\sin \theta)^2 + 2\right] = 0$$ This equation can be solved numerically for \(\theta\). We can then plug these values of \(\theta\) back into the expression for \(f(\cos \theta + 1, \sin \theta)\) to obtain the maximum and minimum values on the boundary. Alternatively, you may use the method of Lagrange multipliers for this task.

Step 6: Compare the values of the function at critical and boundary points

Compare the values of the function at the critical point and the boundary points to determine the absolute maximum and minimum values of \(f(x, y)\) on the set \(R\). Choose the smallest value as the absolute minimum and the largest value as the absolute maximum.

Key Concepts

Critical Points

Critical points are places within the domain of a function where the derivative(s) are zero or undefined. These points can potentially be where the function achieves local maxima, local minima, or saddle points. In the context of multivariable functions, finding critical points involves setting the gradient (a vector of partial derivatives) of the function equal to zero and solving for the variables. This results in a system of equations.

• The gradient of a function \( f(x, y) \) is denoted as \( abla f = (\frac{\partial f}{\partial x}, \frac{\partial f}{\partial y}) \).
• The critical points occur where \( abla f = 0 \).

For example, given the function \( f(x, y) = 2x^2 - 4x + 3y^2 + 2 \), partial derivatives with respect to \( x \) and \( y \) are needed to find the gradient. Solving the system of equations resulting from setting \( \frac{\partial f}{\partial x} = 0 \) and \( \frac{\partial f}{\partial y} = 0 \) helps in locating these potential extrema. In our example, this yields the critical point \((1, 0)\).

Boundary Analysis

Boundary analysis is the process of evaluating a function along the boundary of its domain to find extrema, especially when the domain is restricted. In optimization problems, you need to examine not just the interior of a domain but also its edges or boundaries.

• For a region \( R \), the boundary is where the constraints of \( R \) are equalities, such as \((x-1)^2 + y^2 = 1\) in this problem.
• Substituting boundary constraints into the function changes it into a single-variable problem, which can then be differentiated.

In the example given, the boundary is a circle, which can be expressed as parametric equations using trigonometric identities; \( x-1 \) becomes \( \cos \theta \), and \( y \) becomes \( \sin \theta \). Evaluating the function with these substitutions allows you to explore values of \( \theta \) that maximize or minimize the function along the circular boundary.

Lagrange Multipliers

The method of Lagrange multipliers is a powerful technique used in calculus to find local maxima and minima of a function subject to equality constraints. When you have a function \( f(x, y) \) that you wish to optimize subject to a constraint \( g(x, y) = 0 \), the method introduces an auxiliary variable (the Lagrange multiplier).

• Form the Lagrangian: \( \mathcal{L}(x, y, \lambda) = f(x, y) + \lambda g(x, y) \).
• Compute partial derivatives of \( \mathcal{L} \) with respect to each variable, including \( \lambda \).
• Set each derivative to zero and solve the system of equations.

This method effectively converts a constrained problem into a system of equations, potentially offering solutions that are not apparent through examining the function independently or through boundary analysis alone. While it was referenced as an alternative in this problem, it wasn't explicitly detailed in the solution. However, it offers a systematic approach particularly useful for handling complex constraints.

Partial Derivatives

Partial derivatives are fundamental to understanding multivariable calculus. They give the rate of change of a function with respect to one variable, while keeping the other variables constant. In finding extrema for multivariable functions, partial derivatives help build the gradient.

• \( \frac{\partial f}{\partial x} \): Measures how \( f \) changes as \( x \) changes, keeping \( y \) constant.
• \( \frac{\partial f}{\partial y} \): Measures how \( f \) changes as \( y \) changes, with \( x \) constant.

In the exercise, the given function's partial derivatives were calculated as \( 4x - 4 \) and \( 6y \), respectively. Solving \( \frac{\partial f}{\partial x} = 0 \) and \( \frac{\partial f}{\partial y} = 0 \) led to the critical point. Partial derivatives are essential for understanding the landscape of the function and are foundational for more complex calculus techniques, such as using gradients or Lagrange multipliers.