Question

Absolute maximum and minimum values Find the absolute maximum and minimum values of the following functions over the given regions \(R\). Use Lagrange multipliers to check for extreme points on the boundary. $$f(x, y)=x^{2}+4 y^{2}+1 ; R=\left\\{(x, y): x^{2}+4 y^{2} \leq 1\right\\}$$

Short answer

Answer: The absolute maximum value of the function is 2, and the absolute minimum value is 1.

Step-by-step solution

Find the partial derivatives of the function

To find the critical points, we have to find the partial derivatives of the function with respect to \(x\) and \(y\). Let's find the partial derivatives: $$\frac{\partial f}{\partial x} = 2x$$ $$\frac{\partial f}{\partial y} = 8y$$

Find the critical points by setting the partial derivatives to zero

Now we have to set the partial derivatives to zero and solve for \(x\) and \(y\): $$2x = 0 \Rightarrow x = 0$$ $$8y = 0 \Rightarrow y = 0$$ Thus, the only critical point of the function is at \((x, y) = (0, 0)\), which is within the region \(R\).

Evaluate the function at the critical point

Now, let's evaluate the function at the critical point: $$f(0, 0) = 0^2 + 4 \cdot 0^2 + 1 = 1$$

Use the Lagrange multiplier method to find extreme points on the boundary

Let's use the Lagrange multiplier method to find the extreme points on the boundary. We have the constraint given by the inequality: $$x^2 + 4y^2 \leq 1$$ We can define a new function \(g(x, y) = x^2 + 4y^2 - 1\). The extreme points on the boundary satisfy: $$\nabla f = \lambda \nabla g$$ We have: $$\nabla f = \langle 2x, 8y \rangle$$ $$\nabla g = \langle 2x, 8y \rangle$$ So, the equations to be solved are: $$2x = \lambda (2x)$$ $$8y = \lambda (8y)$$ From the equation \(2x = \lambda (2x)\), we observe that either \(x = 0\) or \(\lambda = 1\). Similarly, from the equation \(8y = \lambda (8y)\), either \(y = 0\) or \(\lambda = 1\). By plugging these values of \(x\) and \(y\) depending on the value of \(\lambda\) into the constraint \(g(x, y) = 0\), we obtain the four points on the boundary of the region \(R\): \((0, \pm \frac{1}{2})\) and \((\pm 1, 0)\).

Evaluate the function at the boundary points

Let's evaluate the function at the boundary points: $$f(0, \pm \frac{1}{2}) = 0^2 + 4 \cdot (\pm\frac{1}{2})^2 + 1 = 2$$ $$f(\pm 1, 0) = (\pm 1)^2 + 4 \cdot 0^2 + 1 = 2$$ Comparing the values of the function at thecritical point \((1)\) and at the boundary points \((2)\), we can see that the absolute minimum value is \(1\) (at \((0, 0)\)), and the absolute maximum value is \(2\) (at \((0, \pm \frac{1}{2})\) and \((\pm 1, 0)\)).

Key concepts

absolute maximum and minimum

In optimization problems, we often seek the absolute maximum and minimum values of a function within a given region. These are the highest and lowest values that the function attains on a specified domain. The absolute maximum is the greatest value, while the absolute minimum is the smallest value. These extreme values can occur at points inside the region as well as on its boundary.

For the function given in the problem, the region over which we need to find these values is defined by the constraint \(x^2 + 4y^2 \leq 1\). By evaluating the function at both the critical points and the boundary, we determined that the absolute minimum of the function is \(1\), occurring at the point \((0, 0)\). The absolute maximum of the function is \(2\), found at the points \((0, \pm \frac{1}{2})\) and \((\pm 1, 0)\). These calculations demonstrate how maximum and minimum values can occur at different locations within the constrained region.

partial derivatives

Partial derivatives are a cornerstone in the study of multivariable calculus, helping us understand how a function changes as its input variables change. They represent the rate of change of a function with respect to one of its variables, while all other variables are held constant.

In the given exercise, to find critical points, partial derivatives of the function \(f(x, y) = x^2 + 4y^2 + 1\) with respect to \(x\) and \(y\) were calculated:

• \(\frac{\partial f}{\partial x} = 2x\)
• \(\frac{\partial f}{\partial y} = 8y\)

By setting these partial derivatives to zero, we identified the points where the function does not increase or decrease in any direction, which are useful in locating potential maximum, minimum, or saddle points of the function.

critical points

Critical points occur where the gradient (vector of partial derivatives) of a function is zero. These points are where a function's rate of change is stationary, neither increasing nor decreasing, which potentially indicates a local max, min, or saddle point.

In the context of the exercise, the partial derivatives were set to zero to find these points:

• For \(\frac{\partial f}{\partial x} = 2x\), setting \(2x = 0\) gives \(x = 0\).
• For \(\frac{\partial f}{\partial y} = 8y\), setting \(8y = 0\) gives \(y = 0\).

Thus, the critical point on our domain is at \((0,0)\), where the evaluated function value is \(1\). In this exercise, this critical point is the location of the absolute minimum value.

extreme points on the boundary

Finding extreme points on the boundary involves using the Lagrange multiplier method, a technique that allows us to handle constraints effectively when seeking max or min values of a function. This method is especially useful because boundary points often do not align with gradients being zero, so a different technique is required.

We employ Lagrange Multipliers by setting up the Lagrangian function with a constraint and comparing gradients:

• The function \(f(x, y) = x^2 + 4y^2 + 1\) has the constraint \(g(x, y) = x^2 + 4y^2 - 1 = 0\), representing the boundary of our region.
• Setting \(abla f = \lambda abla g\), and solving the system of equations, gives potential points on the boundary: \((0, \pm \frac{1}{2})\) and \((\pm 1, 0)\).

Evaluating the function at these points shows they are extreme, with better values than even some points inside the boundary.