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. $$f(x, y)=(x-1)^{2}+(y+1)^{2} ; R=\left\\{(x, y): x^{2}+y^{2} \leq 4\right\\}$$

Short answer

Answer: The absolute maximum value of the function is 4 and the absolute minimum value is 0.

Step-by-step solution

Find critical points of the function inside the region

To find the critical points of the function inside the region, we need to find where the gradient of the function is equal to 0. We will do this by taking partial derivatives of the function with respect to \(x\) and \(y\) and then solving for the points where both partial derivatives are equal to 0. $$f_x = \frac{\partial}{\partial x}\left((x-1)^2 + (y+1)^2\right) = 2(x-1)$$ $$f_y = \frac{\partial}{\partial y}\left((x-1)^2 + (y+1)^2\right) = 2(y+1)$$ Now, solve both partial derivatives equal to 0: $$2(x-1) = 0 \Rightarrow x = 1$$ $$2(y+1) = 0 \Rightarrow y = -1$$ So the critical point inside the region is \((1, -1)\).

Find the value of the function at the critical point inside the region

Now we will find the value of the function at the critical point we found in the previous step: $$f(1, -1) = (1-1)^2 + (-1+1)^2 = 0$$

Apply Lagrange multipliers on the boundary

Next, we will apply Lagrange multipliers to find extreme points on the boundary: Let \(g(x, y) = x^2 + y^2\), being the function that represents the constraint. Then \(\nabla f = \lambda \nabla g\), or equivalently $$f_x = \lambda g_x$$ $$f_y = \lambda g_y$$ Substituting the expressions of partial derivatives, we have $$2(x - 1) = 2\lambda x$$ $$2(y + 1) = 2\lambda y$$ From these equations, we get: $$x - 1 = \lambda x$$ $$y + 1 = \lambda y$$

Solve the system of equations

Now, solve the system of equations: $$\lambda = \frac{x-1}{x}$$ $$\lambda = \frac{y+1}{y}$$ Multiplying the denominators we get: $$y(x-1)=x(y+1)$$ This simplifies to: $$xy-x=yx+y$$ \(x-y=0\) or \(y=x\) Now we will substitute \(y = x\) back into the constraint equation \(g(x, y) = x^2 + y^2 = 4\): $$x^2 + x^2 = 4$$ $$2x^2 = 4$$ $$x^2 = 2$$ Hence, \(x = \pm \sqrt{2}\) and \(y = \pm \sqrt{2}\).

Evaluate the function at the extreme points

Now we will evaluate the function at the extreme points we found: $$f(\sqrt{2}, \sqrt{2}) = (\sqrt{2} - 1)^2 + (\sqrt{2} + 1)^2 = 1 + 3 = 4$$ $$f(-\sqrt{2}, -\sqrt{2}) = (-\sqrt{2} - 1)^2 + (-\sqrt{2} +1)^2 = 1 + 3 = 4$$ So the minimum value of the function over the given region is \(f(1, -1) = 0\), and the maximum value is \(f(\sqrt{2}, \sqrt{2}) = f(-\sqrt{2}, -\sqrt{2}) = 4\).

Key concepts

Critical Points

In the context of functions, critical points are where the function's derivative equals zero or is undefined. They are vital because they can indicate where local maxima or minima occur, or where the function changes behavior. For multivariable functions like the one in this exercise, critical points occur where all partial derivatives are zero.

To find critical points within a region, calculate the partial derivatives of the function with respect to each variable. Set each derivative to zero and solve the resulting equations simultaneously.

• For the function \(f(x, y)=(x-1)^2+(y+1)^2\), the partial derivatives are:

\(f_x=2(x-1)\) and \(f_y=2(y+1)\).
Solving for zero, you find the critical point at \((1, -1)\). This means that at \(x=1\), and \(y=-1\), the gradient of the function is zero, indicating a possible extreme point inside the region.

Gradient

The gradient of a function refers to a vector formed by its partial derivatives. It gives the direction of the steepest ascent of the function. In a multivariable setting, the gradient is represented as \(abla f = \left(\frac{\partial f}{\partial x}, \frac{\partial f}{\partial y}\right)\).

When the gradient is zero at a point, it suggests we've reached a critical point, which can be a point of local maxima, minima, or a saddle point.

• For \(f(x, y)=(x-1)^2 + (y+1)^2\), the gradient is \(abla f = (2(x-1), 2(y+1))\).

At critical points like \((1, -1)\), the gradient is \((0, 0)\).

This tells us the function is neither increasing nor decreasing, and thus we're on a horizontal tangent plane in the function's 3D "landscape." Investigating these critical points helps determine the function's behavior at those locations.

Partial Derivatives

Partial derivatives are foundational when dealing with functions of several variables. They represent the rate of change of the function concerning one variable while keeping others constant. For a function \(f(x, y)\), the partial derivative with respect to \(x\) is obtained by differentiating as if \(y\) were a constant and vice versa.

In this exercise:

• \(f_x = \frac{\partial}{\partial x}((x-1)^2 + (y+1)^2) = 2(x-1)\) measures change along the \(x\)-axis.
• \(f_y = \frac{\partial}{\partial y}((x-1)^2 + (y+1)^2) = 2(y+1)\) measures change along the \(y\)-axis.

These derivatives are essential for finding critical points and analyzing the behavior of the function in various regions.

By understanding partial derivatives, learners grasp how each variable influences the function, a critical skill in multivariable calculus.

Extreme Points

Extreme points, or extrema, refer to points where a function takes on its maximum or minimum values. In multivariable calculus, finding extrema involves checking critical points and using methods like Lagrange multipliers on boundaries.

For internal points, set the gradient to zero to find candidates for extrema.

At boundaries, as in this exercise with constraint \(x^2 + y^2 = 4\), apply Lagrange multipliers. Here's how it's done:

• Introduce a constraint function \(g(x, y) = x^2 + y^2 - 4\).
• The Lagrange condition \(abla f = \lambda abla g\) gives a system of equations to solve for \(\lambda\) and critical points.
• Solving gives \((x, y) = (\sqrt{2}, \sqrt{2})\) and \((-\sqrt{2}, -\sqrt{2})\).

These points on the boundary were evaluated to find possible maximum values.

Ultimately, identifying extreme points relies on proper use of the derivative strategies and techniques like Lagrange multipliers.