Applications of function of two variables
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Applications of Extrema of Functions of Two Variables
Purpose
The purpose of this lab is to give you experience in applying calculus techniques relating to finding extrema of functions of two variables.
Structure
There are three problems, each of which has a background discussion, an illustrative example, and an exercise for you to do. Most of the basic theoretical background you will need has been covered in Lab 3, but there will be a little repetition here for completeness. Most of the Maple procedures you will need have been covered in previous labs, but there will be a few new ones introduced below.
Problem 1: Finding Extrema
Background
For some simple functions of two variables, it is not difficult to determine their relative extrema by first finding the critical points and then applying the Second Partials Test (SPT) to discriminate among relative maxima, relative minima, and saddle points. Recall that if the function f(x,y) is differentiable, then (x0,y0) is a critical point if . With D=fxx(x0,y0)fyy(x0,y0)-[fxy(x0,y0)]2, the SPT is as follows:
If D>0 and fxx(x0,y0)>0, then f has a relative minimum at (x0, y0). If D>0 and fxx(x0,y0)<0, then f has a relative maximum at (x0, y0). If D<0, then (x0, y0) is a saddle point of f. If D=0, then the test is inconclusive.
Many applications described by functions of two variables can be studied by purely analytical means, but computer software such as Maple might be of substantial help in gaining further insight or obtaining information that cannot be found analytically. In Lab 3, you used Maple tools such as diff, solve, fsolve for finding critical points and extrema and applying the SPT. In this lab, we will encourage the use of more powerful Maple tools, such as the grad command mentioned in Lab 3, in combination with tools for visualization such as plot and plot3d.
Example
To find the relative extrema of f(x,y) = -x3 + 4xy - 2y2 + 1, first find the critical points by determining where fx(x,y)=fy(x,y)=0. Solving these equations analytically leads to the two critical points (0,0) and (4/3,4/3). Applying the SPT to the critical points, you find that (0,0) corresponds to a saddle point of f, whereas f has a relative maximum at (4/3,4/3).
Purpose
The purpose of this lab is to give you experience in applying calculus techniques relating to finding extrema of functions of two variables.
Structure
There are three problems, each of which has a background discussion, an illustrative example, and an exercise for you to do. Most of the basic theoretical background you will need has been covered in Lab 3, but there will be a little repetition here for completeness. Most of the Maple procedures you will need have been covered in previous labs, but there will be a few new ones introduced below.
Problem 1: Finding Extrema
Background
For some simple functions of two variables, it is not difficult to determine their relative extrema by first finding the critical points and then applying the Second Partials Test (SPT) to discriminate among relative maxima, relative minima, and saddle points. Recall that if the function f(x,y) is differentiable, then (x0,y0) is a critical point if . With D=fxx(x0,y0)fyy(x0,y0)-[fxy(x0,y0)]2, the SPT is as follows:
If D>0 and fxx(x0,y0)>0, then f has a relative minimum at (x0, y0). If D>0 and fxx(x0,y0)<0, then f has a relative maximum at (x0, y0). If D<0, then (x0, y0) is a saddle point of f. If D=0, then the test is inconclusive.
Many applications described by functions of two variables can be studied by purely analytical means, but computer software such as Maple might be of substantial help in gaining further insight or obtaining information that cannot be found analytically. In Lab 3, you used Maple tools such as diff, solve, fsolve for finding critical points and extrema and applying the SPT. In this lab, we will encourage the use of more powerful Maple tools, such as the grad command mentioned in Lab 3, in combination with tools for visualization such as plot and plot3d.
Example
To find the relative extrema of f(x,y) = -x3 + 4xy - 2y2 + 1, first find the critical points by determining where fx(x,y)=fy(x,y)=0. Solving these equations analytically leads to the two critical points (0,0) and (4/3,4/3). Applying the SPT to the critical points, you find that (0,0) corresponds to a saddle point of f, whereas f has a relative maximum at (4/3,4/3).
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