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Find the global maximum and minimum of the function on \(-1 \leq x \leq 1,-1 \leq y \leq 1,\) and say whether it occurs on the boundary of the square. [Hint: Use graphs.] $$z=x^{2}-y^{2}$$

Short Answer

Expert verified
Global max of 1 at (1,0) and (-1,0); global min of -1 at (0,1) and (0,-1); on boundary.

Step by step solution

01

Identify the Level Surfaces

The function given is of the form \(z = x^2 - y^2\). Notice that this represents a hyperbolic paraboloid, which is saddle-shaped. The level surfaces are sets of points \((x, y)\) that satisfy \(x^2 - y^2 = c\) for various values of \(c\). The function's behavior changes along the axes, so dissecting along these lines helps in understanding the surface better.
02

Check Horizontal and Vertical Axes

On the line \(y = 0\), \(z = x^2\), which is a parabola opening upwards. On the line \(x = 0\), \(z = -y^2\), which is a parabola opening downwards. These observations indicate possible plane intersections at \((x, 0)\) and \((0, y)\). Being a saddle point structure, it tells us there are likely boundary points worth checking.
03

Evaluate Function at Vertex and Critical Points

Check \(z = x^2 - y^2\) at the center point \((0,0)\). Here, \(z = 0 - 0 = 0\). However, it isn't necessarily the extremum because of the nature of a saddle point. Analyze the boundary lines and intersections.
04

Evaluate Function at Boundary Lines

The boundary consists of the lines at the edges of the square. Evaluate \(z\) at these lines.- On \(x = 1\) or \(x = -1\), \(z = 1 - y^2\), maximum at \(y = 0\) gives \(z = 1\), and minimum at \(y = \pm 1\) gives \(z = 0\).- On \(y = 1\) or \(y = -1\), \(z = x^2 - 1\), maximum at \(x = \pm 1\) gives \(z = 0\), and minimum at \(x = 0\) gives \(z = -1\).
05

Determine Global Extremes and Their Locations

The global maximum can be found by comparing these values. The maximum value is \(z = 1\), occurring at \((x, y) = (1, 0)\) and \((-1, 0)\). The global minimum is \(z = -1\), which occurs at \((x, y) = (0, 1)\) and \((0, -1)\). All extrema occur on the boundary of the domain.

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Hyperbolic Paraboloid
The hyperbolic paraboloid is a fascinating surface in 3D geometry, often described as a saddle surface because of its unique shape. Visualize the surface as a smooth, curving saddle, which dips and rises at various points. In the context of the given problem, the function \(z = x^2 - y^2\) is represented by a hyperbolic paraboloid, showcasing a distinctive form that dips and rises symmetrically.

- At any constant level \(c\), the equation \(x^2 - y^2 = c\) describes a hyperbola on the \(xy\)-plane.- This form means that for every positive \(c\), the function looks like a hyperbola opening along one axis, and for negative \(c\), it shifts direction.- This saddle-like shape implies an inherent asymmetry—demonstrating its versatility in engineering and design examples, such as roofs and bridges.Encountering it in mathematical problems allows students to understand changes in curvature across a surface, making it central for analyzing properties like minimal surfaces.
Level Surfaces
Level surfaces are crucial when visualizing 3D shapes from 2D equations. Often described as contours in a mountain range, they represent cross-sections at various heights. For the function \(z = x^2 - y^2\), level surfaces are determined by setting \(z = c\), where \(c\) is a constant. Here's how to interpret them:

- **Planes of constant value**: Consider lines like \(x = 0\) or \(y = 0\), producing planes such as \(z = x^2\) or \(z = -y^2\).- By solving \(z - c = x^2 - y^2\) for different \(c\) values, students can visualize these sections as hyperbolas.- Understanding their intersections with the axes can help predict the global behavior of the function.By visualizing level surfaces, learners grasp how functions change within specific regions, assisting in predicting critical points or extrema.
Saddle Point
The saddle point is a pivotal concept in calculus, symbolizing a point where a function changes direction but does not express a local extremum. For the hyperbolic paraboloid \(z = x^2 - y^2\), the saddle point is at \((x, y) = (0, 0)\), where \(z = 0\).

- **Characteristics**: At points like \((0,0)\), the curvature directs upwards along one axis and downwards along another.- **Analysis**: Despite seeming like a minimum or maximum, it is neither due to opposing curvatures at the point.The saddle point exemplifies how changes occur symmetrically around a specific location, preparing students to analyze less straightforward scenarios in multi-variable calculus.
Boundary Evaluation
Boundary evaluation involves assessing a function's behavior at the edges of a given domain. Useful for determining global extrema, especially when dealing with defined constraints or boundaries. For our function \(z = x^2 - y^2\) within the square bounded by \(-1 \leq x \leq 1\) and \(-1 \leq y \leq 1\), boundary conditions offer insights into where maximum and minimum values occur.

- **Line Checking**: By evaluating \(z\) on each side of the boundary square, one can identify values such as \(z = 0\) or \(z = -1\).- **Comparison**: This involves comparing values at the boundary against those at the saddle point.In this case, the global maximum and minimum values occur precisely at these boundaries, highlighting boundary evaluation's role in optimizing solutions in various real-world problems.

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Most popular questions from this chapter

The behavior of a function can be complicated near a critical point where \(D=0 .\) Suppose that $$f(x, y)=x^{3}-3 x y^{2}$$ Show that there is one critical point at (0,0) and that \(D=0\) there. Show that the contour for \(f(x, y)=0\) consists of three lines intersecting at the origin and that these lines divide the plane into six regions around the origin where \(f\) alternates from positive to negative. Sketch a contour diagram for \(f\) near \((0,0) .\) The graph of this function is called a monkey saddle.

Give an example of: A function \(f(x, y)\) and a region \(R\) such that the maximum value of \(f\) on \(R\) is on the boundary of \(R\).

Let \(h(x, y)=f(x)+g(y) .\) Show that \(h\) has a critical point at \((a, b)\) if \(f^{\prime}(a)=g^{\prime}(b)=0,\) and, assuming \(f^{\prime \prime}(a) \neq 0\) and \(g^{\prime \prime}(b) \neq 0,\) it is a local maximum or minimum when \(f^{\prime \prime}(a)\) and \(g^{\prime \prime}(b)\) have the same sign and a saddle point when they have opposite signs.

Design a rectangular milk carton box of width \(w\), length \(l,\) and height \(h\) which holds \(512 \mathrm{cm}^{3}\) of milk. The sides of the box cost 1 cent \(/ \mathrm{cm}^{2}\) and the top and bottom cost 2 cent \(/ \mathrm{cm}^{2} .\) Find the dimensions of the box that minimize the total cost of materials used.

At the point \((1,3),\) suppose that \(f_{x}=f_{y}=0\) and \(f_{x x}>0, f_{y y}>0, f_{x y}=0\) (a) What can you conclude about the behavior of the function near the point (1,3)\(?\) (b) Sketch a possible contour diagram.

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