Problem 15

Question

The functions are defined on the rectangular domain $$ D=\\{(x, y):-1 \leq x \leq 1,-1 \leq y \leq 1\\} $$ Find the absolute maxima and minima of $f$ on $D$. $f(x, y)=x^{2}-y^{2}$

Step-by-Step Solution

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Answer

The absolute maximum is 1, and the absolute minimum is 0.

1Step 1: Determine the Critical Points

First, find the critical points by taking the partial derivatives of the function and setting them equal to zero. The partial derivatives are:\[ f_x = 2x, \quad f_y = -2y. \]Setting $ f_x = 0 $ and $ f_y = 0 $ gives $ x = 0 $ and $ y = 0 $. Therefore, the critical point is $(0, 0)$.

2Step 2: Evaluate at Critical Point

Evaluate $ f(x, y) $ at the critical point $(0, 0)$:\[ f(0, 0) = 0^2 - 0^2 = 0. \]This is one potential candidate for the extrema.

3Step 3: Evaluate on the Boundary

Next, evaluate the function along the boundaries of $ D $. This is done by fixing the values of $ x $ and $ y $ at the edges of the domain.1. For the boundary $ x = 1 $: $ f(1, y) = 1^2 - y^2 = 1 - y^2 $. - Evaluate at $ y = -1 $ and $ y = 1 $, which gives $ f(1, -1) = 0 $ and $ f(1, 1) = 0 $.2. For the boundary $ x = -1 $: $ f(-1, y) = (-1)^2 - y^2 = 1 - y^2 $. - Evaluate at $ y = -1 $ and $ y = 1 $, which gives $ f(-1, -1) = 0 $ and $ f(-1, 1) = 0 $.3. For the boundary $ y = 1 $: $ f(x, 1) = x^2 - 1 $. - Evaluate at $ x = -1 $ and $ x = 1 $, which gives $ f(-1, 1) = 0 $ and $ f(1, 1) = 0 $.4. For the boundary $ y = -1 $: $ f(x, -1) = x^2 - 1 $. - Evaluate at $ x = -1 $ and $ x = 1 $, which gives $ f(-1, -1) = 0 $ and $ f(1, -1) = 0 $.

4Step 4: Consider Interior Evaluations

For fixed boundaries where $ x $ or $ y $ are within the interval $[-1, 1]$, we already evaluate this effectively by ensuring endpoints on boundary checks are sufficient.For example, for $ y = 0 $, the function simplifies to $ f(x, 0) = x^2 $ which takes a maximum at the boundary where $ |x| = 1 $ leading to value 1.

5Step 5: Determine Absolute Maxima and Minima

Compare all the evaluations:- Critical point $ f(0, 0) = 0 $- Boundaries lead to $ f(x, y) = 1 $ or $ 0 $ for given input values.Thus the absolute maximum value is 1 and minimum is 0 on the rectangular domain $ D $.

Key Concepts

Critical PointsPartial DerivativesBoundary EvaluationRectangular Domain

Critical Points

Critical points play a crucial role in determining where a function might reach a local maximum or minimum value. In multivariable calculus, to find these points, we calculate the partial derivatives of a function. Partial derivatives are like regular derivatives but hold other variables constant. Once calculated, we set these derivatives equal to zero to find the critical points.

For the function, $ f(x, y) = x^2 - y^2 $, partial derivatives are $ f_x = 2x $ and $ f_y = -2y $.
Setting $ f_x = 0 $ gives $ 2x = 0 $, or $ x = 0 $.
Setting $ f_y = 0 $ gives $ -2y = 0 $, or $ y = 0 $.

This means our critical point is $(0, 0)$. At this point, the changes in $ x $ and $ y $ provide no slope, indicating a potential extremum or saddle point.

Partial Derivatives

Partial derivatives are essential tools in multivariable calculus for understanding how a function changes along a particular direction. They measure the rate of change of the function with respect to one variable while keeping the other variables constant.

With $ f(x, y) = x^2 - y^2 $, if you keep $ y $ constant, you get the partial derivative with respect to $ x $, $ f_x = 2x $.
If you keep $ x $ constant, the partial derivative with respect to $ y $, $ f_y = -2y $, is obtained.

By setting these partial derivatives to zero, we can identify critical points, providing insight into the function's behavior and potential extremum locations.

Boundary Evaluation

Evaluating along the boundaries of a set domain is necessary to ensure you capture all extrema, as they might not occur strictly at critical points.

The boundaries of the rectangular domain $ D $ involve setting either $ x = 1 $, $ x = -1 $, $ y = 1 $, or $ y = -1 $.
For example, along $ x = 1 $, the function becomes $ f(1, y) = 1 - y^2 $, which simplifies evaluations at $ y = -1 $ and $ y = 1 $ to zero.

Checking every domain edge ensures any maxima or minima present at these limits are not overlooked. In our case, evaluating the function on these boundaries also supports identifying where the extremum lies.

Rectangular Domain

A rectangular domain describes a specific area within the function's variables are bounded. This domain plays a pivotal role in multivariable optimization problems.

In this exercise, the domain $ D = \{(x, y) : -1 \leq x \leq 1, -1 \leq y \leq 1\} $ indicates the region over which we seek extrema.
This domain allows us to comprehend where both critical points and boundaries need evaluation.

Ensuring we examine both interior and edge of these domains ensures that any maximum or minimum values won't be missed. This is critical, as functions can behave very differently within and on the outside edges of these constraints.

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