If your solution to a stated problem does not match the given solution, you should self-critique per instructions at  

http://vhcc2.vhcc.edu/dsmith/geninfo/labrynth_created_fall_05/levl1_22/levl2_81/file3_259.htm.

Your solution, attempt at solution: If you are unable to attempt a solution, give a phrase-by-phrase interpretation of the problem along with a statement of what you do or do not understand about it. This response should be given, based on the work you did in completing the assignment, before you look at the given solution.

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Question: `qQuery problem 7.5.10 extrema of x^2+6xy+10y^2-4y+4

List the relative extrema and the saddle points of the function and tell which is which, and how you obtained each.

Your solution:

Confidence Assessment:

Given Solution:

`a fx = 2x + 6y ; fy = 6x + 20 y - 4, where fx and fy mean x and y partial derivatives of f.

fxx is the x derivative of fx and is therefore 2

fyy is the y derivative of fy and is therefore 20

fxy is the y derivative of fx and is therefore 6.  Note that fyx is the x derivative of fy and is therefore 6.  fxy and fyx are always the same, provided the derivatives exist.

fx = 0 and fy = 0 if

2x + 6y = 0 and

6x + 20y - 4 = 0.

This is a system of two equations in two unknowns.  You can solve the system by any of several methods.  Multiplying the first equation by -3 and adding the resulting equations, for example, gives you 2 y = 4 so that y = 2.  Substituting y = 2 into the first equation gives you 2x + 6 * 2 = 0, which has solution x = 6.

Whatever method you use, the solution of this system is x = -6, y = 2.

To determine the nature of the critical point at (-6, 2) you have to look at fxx, fyy and fxy.  We have

fxx = 2

fyy = 20

fxy = 6.

This is a system of two equations in two unknowns.  You can solve the system by any of several methods.  Multiplying the first equation by -3 and adding the resulting equations, for example, gives you 2 y = 4 so that y = 2.  Substituting y = 2 into the first equation gives you 2x + 6 * 2 = 0, which has solution x = -6.

Whatever method you use, the solution of this system is x = -6, y = 2.

To determine the nature of the critical point at (-6, 2) you have to evaluate the quantity fxx * fyy - 4 fxy^2.

We have

fxx = 2

fyy = 20

fxy = 6.

So fxx * fyy - fxy^2 = 2 *  20 - 6^2 = 4.

This quantity is positive, so you have either a maximum or a minimum.

Since fxx and fyy are both positive the graph is concave upward in all directions and the point is a minimum.

The coordinates of the point are (-6, 2, (-6)^2+6 * (-6) * 2 + 10* 2^2 - 4 * 2 + 4  ) = (-6, 2, 0)

Self-critique (if necessary):

Self-critique Rating:

Question: `qQuery problem 7.5.28 extrema of x^3+y^3 -3x^2+6y^2+3x+12y+7

List the relative extrema and the saddle points of the function and tell which is which, and how you obtained each.

Your solution:

Confidence Assessment:

Given Solution:

`a We have fx = 3 x^2 + 6 x + 3 and fy = 3 y^2 + 12 y + 12.

Factoring we get

fx = 3 ( x^2 - 2x + 1) = 3 ( x-1)^2 and

fy = 3 ( y^2 + 4y + 4) = 3 ( y+2)^2.

So

fx = 0 when 3(x-1)^2 = 0, or x = 1 and

fy = 0 when 3(y+2)^2 = 0 or y = -2.

We get

fxx = 6 x - 6 (the x derivative of fx)

fyy = 6 y + 12 (the y derivative of fy) and

fxy = 0 (the y derivative of fx, which is also equal to fyx, the x derivative of fy)

At the critical point x = 1, y = -2 we get fxx = fyy = 0.  So the test for max, min and critical point gives us fxx*fyy - fxy^2 = 0, which is inconclusive--it tells us nothing about max, min or saddle point.

Self-critique (if necessary):

Self-critique Rating: