Question

Find the absolute maximum and minimum values of the following functions on the given set \(R\). $$\begin{aligned} &f(x, y)=-2 x^{2}+4 x-3 y^{2}-6 y-1\\\ &R=\left\\{(x, y):(x-1)^{2}+(y+1)^{2} \leq 1\right\\} \end{aligned}$$

Short answer

Answer: The absolute maximum value of the function on the given set R is 3.

Step-by-step solution

Find critical points inside the region R

First, compute the partial derivatives with respect to x and y: $$\begin{aligned} \frac{\partial f}{\partial x} &= -4x + 4 \\ \frac{\partial f}{\partial y} &= -6y - 6 \\ \end{aligned}$$ Set these partial derivatives to 0 and solve for x and y: $$\begin{aligned} -4x + 4 &= 0 \Rightarrow x=1 \\ -6y - 6 &= 0 \Rightarrow y=-1 \\ \end{aligned}$$ So there is one critical point inside R: \((1, -1)\).

Find the maximum and minimum values on the boundary of the region R

To find the maximum and minimum values on the boundary, we use the method of Lagrange multipliers. We construct the following auxiliary function: $$ G(x, y, \lambda) = -2 x^{2}+4 x-3 y^{2}-6 y-1 + \lambda \left((x-1)^{2}+(y+1)^{2}-1\right) $$ Compute the derivatives with respect to x, y, and \(\lambda\): $$\begin{aligned} \frac{\partial G}{\partial x} &= -4x + 4 + 2\lambda(x-1) \\ \frac{\partial G}{\partial y} &= -6y - 6 + 2\lambda(y+1) \\ \frac{\partial G}{\partial \lambda} &= (x-1)^{2}+(y+1)^{2}-1 \\ \end{aligned}$$ Set these to 0 and solve for x, y, and \(\lambda\): $$\begin{aligned} -4x + 4 + 2\lambda(x-1) &= 0 \\ -6y - 6 + 2\lambda(y+1) &= 0 \\ (x-1)^{2}+(y+1)^{2}-1 &= 0 \\ \end{aligned}$$ From equations (1) and (2), we get: $$\begin{aligned} \lambda (x-1) &= x-1 \\ \lambda (y+1) &= -3y-3\\ \end{aligned}$$ From the third equation, we know \((x-1)^{2}+(y+1)^{2}=1\). There are two possible cases: Case 1: \(\lambda = 1\). In this case, \(x=1\) and \(y=-1\), which is the critical point we've already found. Case 2: \(\lambda \neq 1\). Since \(\lambda \neq 1\), we can divide the equations by \((1-\lambda)\): $$\begin{aligned} x-1 &= 1 \\ y+1 &= -\frac{3}{2} \\ \end{aligned}$$ So, the point on the boundary is \((2, -5/2)\). However, this point does not satisfy the equation \((x-1)^{2}+(y+1)^{2}=1\), so it is not on the boundary of R. Thus, there are no additional points to consider.

Compare the values of the function at the critical points and the boundary

Compute the value of the function at the critical point (1,-1): $$f(1, -1) = -2(1)^2 + 4(1) -3(-1)^2 - 6(-1) -1 = 3$$ Since there were no valid points on the boundary, the absolute maximum value of the function on the given set R is 3, and the absolute minimum value does not exist.

Key concepts

Lagrange multipliers

Lagrange multipliers are a strategic mathematical method used for finding the local maxima and minima of functions subject to equality constraints. Suppose you have a function, like our exercise's function, which you want to optimize within a certain boundary. With Lagrange multipliers, you introduce a new variable called the Lagrange multiplier, generally denoted by \( \lambda \), to create an auxiliary function. This auxiliary function combines the original function and the constraint. The equation typically looks like this: \[ G(x, y, \lambda) = f(x, y) + \lambda \, g(x, y) \] where \( f(x, y) \) is the function you want to optimize, and \( g(x, y) = 0 \) represents the constraint equation.To find the extrema (maximum or minimum points), you take partial derivatives of this auxiliary function concerning the original variables and \( \lambda \). Setting these partial derivatives to zero gives you a system of equations. Solving this system will help identify potential optimal points, both within the region and along the edges of the boundary.In our exercise, we constructed an auxiliary function with \( G(x, y, \lambda) = -2x^2 + 4x - 3y^2 - 6y - 1 + \lambda((x-1)^2 + (y+1)^2 - 1) \), and then computed derivatives with respect to \( x \), \( y \), and \( \lambda \) to find the points satisfying the boundary constraint.

partial derivatives

Partial derivatives are crucial tools in calculus when dealing with functions of multiple variables. They let us understand how a function changes as we vary one variable while holding the others constant. In simpler terms, it's like examining the slope of a surface in a particular direction.For any function \( f(x, y) \), the partial derivative with respect to \( x \) (denoted as \( \frac{\partial f}{\partial x} \)), gives us the rate of change of the function as \( x \) changes while \( y \) remains constant. Similarly, the partial derivative with respect to \( y \) tells us how the function changes as \( y \) changes, keeping \( x \) fixed.In our exercise, the function is \( f(x, y) = -2x^2 + 4x - 3y^2 - 6y - 1 \). Taking partial derivatives:

• \( \frac{\partial f}{\partial x} = -4x + 4 \)
• \( \frac{\partial f}{\partial y} = -6y - 6 \)

These derivatives are then set to zero in order to locate critical points. This step is key to finding where our function might reach local maxima or minima.

critical points

Critical points are specific values of \( x \) and \( y \) at which the partial derivatives of a function are zero. These points are vital because they are potential locations where the function could reach its highest or lowest values, often referred to as extrema.To find critical points, you need to solve for where \( \frac{\partial f}{\partial x} = 0 \) and \( \frac{\partial f}{\partial y} = 0 \) simultaneously. These solutions give us the coordinates of critical points. In the provided exercise, setting the partial derivatives to zero:

• \( -4x + 4 = 0 \), simplifies to \( x = 1 \)
• \( -6y - 6 = 0 \), simplifies to \( y = -1 \)

Therefore, the critical point within the set \( R \) is \((1, -1)\). Checking the value of \( f(1, -1) = 3 \) illustrates that this is where the absolute maximum occurs within the defined region, as no other valid boundary information was found in the step-by-step solution. This point is important as it highlights the utility of critical points in optimization problems.