Question

Find the absolute maximum and minimum values of the following functions on the given set \(R\). \(f(x, y)=x^{2}+y^{2}-2 x-2 y ; R\) is the closed set bounded by the triangle with vertices \((0,0),(2,0),\) and (0,2)

Short answer

Answer: The absolute minimum value of the function is -1 at the point (1, 1), while there is no absolute maximum value inside the triangle \(R\), but the function value is 0 at all vertices of the triangle.

Step-by-step solution

Finding critical points

To find the critical points of the function, \(f(x, y)=x^{2}+y^{2}-2 x-2 y\), it is necessary to find the first order partial derivatives of \(f\) with respect to \(x\) and \(y\). The critical points occur when both of theses derivatives equal to zero (as long as the second order partial derivatives exist). The partial derivatives are as follows: \(\frac{\partial f}{\partial x} = 2x - 2\) \(\frac{\partial f}{\partial y} = 2y - 2\) Setting both these equations equal to 0, we have: \(2x - 2 = 0 \Rightarrow x = 1\) \(2y - 2 = 0 \Rightarrow y = 1\) So, we have found the critical point \((1, 1)\).

Evaluating the function at critical point and vertices of the triangle

Now, we need to evaluate the function at the critical point and the vertices (boundary points) of the triangle \(R\): \(f(0,0) = (0)^{2} + (0)^{2} -2(0) - 2(0) = 0\) \(f(2,0) = (2)^{2} + (0)^{2} -2(2) - 2(0) = 0\) \(f(0,2) = (0)^{2} + (2)^{2} -2(0) - 2(2) = 0\) \(f(1,1) = (1)^{2} + (1)^{2} -2(1) -2(1) = -1\)

Determining the absolute maximum and minimum values

Comparing the function values at the vertices of the triangle and the critical point: Absolute min: \(f(x, y) = -1\) at \((1, 1)\) Absolute max: \(f(x, y) = 0\) does not exist inside the triangle \(R\) but the function value is the same at all vertices of the triangle \((0, 0)\), \((2, 0)\), and \((0, 2)\).

Key concepts

Critical Points

When solving for absolute extrema, locating the critical points of the function is crucial. Critical points are where the first partial derivatives of a function are zero or undefined. For a function of two variables like \( f(x, y) \), we find critical points by taking the partial derivatives with respect to each variable and setting them equal to zero.

In this exercise, the function \( f(x, y) = x^2 + y^2 - 2x - 2y \) required the derivatives \( \frac{\partial f}{\partial x} \) and \( \frac{\partial f}{\partial y} \). Solving \( 2x - 2 = 0 \) and \( 2y - 2 = 0 \), we found the critical point at \( (1, 1) \). This point is potentially where the function has local minima or maxima. It's important to note that not all critical points guarantee absolute extrema; further evaluation is necessary.

Partial Derivatives

Partial derivatives are essential when differentiating functions of multiple variables. They help us determine how a function changes as each individual variable changes, keeping the others constant. In optimization tasks, they provide directions for steepest increase or decrease of function values.

For \( f(x, y) = x^2 + y^2 - 2x - 2y \):

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

These partial derivatives suggest how \( f(x, y) \) behaves as \( x \) or \( y \) varies and are set to zero to find critical points. Understanding these derivatives is key in revealing the nature of the surface described by the function.

Boundary Evaluation

Boundary evaluation is crucial in identifying absolute extrema for functions restricted to a specific region. After finding the critical points within the interior of the region, examining function values on the boundary ensures no extrema are missed.

For our exercise, the region \( R \) is defined by a triangle with vertices \((0,0), (2,0), (0,2)\). The function values were computed at these vertices. We checked

• \( f(0, 0) = 0 \)
• \( f(2, 0) = 0 \)
• \( f(0, 2) = 0 \)

All of these values yielded 0, showing that the boundary could have potential maximum value locations. Boundary evaluation helps determine the behavior of \( f(x, y) \) at the edges, ensuring comprehensive exploration of extrema.

Optimization in Calculus

Optimization in calculus involves finding the largest or smallest values a function can take, subject to certain constraints or within a specific domain. This process typically involves several steps: finding critical points, evaluating the function at these points, and assessing boundary values.

In the given task, optimization involved

• Finding critical points by setting partial derivatives equal to zero.
• Evaluating function values at these points and across region boundaries defined by a triangle.
• Comparing all computed values to identify absolute maxima and minima.

Optimization helps us make informed decisions on function behavior. By understanding critical values and regional constraints, students can grasp how calculus facilitates deriving meaningful insights about complex surfaces.