If your solution to 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
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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.
At the end of this document, after the qa problems (which provide you with questions and solutions), there is a series of Questions, Problems and Exercises.
11.5
Question: `q001. Consider f(x,y) = 3x^2 - 5xy + y^2 + 3. Find the critical points, and classify each point as either a relative maximum, relative minimum, or saddle point.
Your solution:
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Given Solution:
f_x = 6 x - 5 y and f_y = 5 x + 2 y. Our critical point therefore occurs when
6x - 5y = 0 and
5x + 2y = 0.
Solving simultaneously we get x = 0, y = 0 so our critical point is (0, 0).
The second derivatives are f_xx = 6, f_yy = 2 and f_xy = -5. f_xx and f_yy are both positive, indicating that if the critical point is not a saddle point, it is a minimum
f_xx * f_yy - f_xy^2 = 6 * 2 - (-5)^2 = -13, which is < 0 and indicates that the critical point is a saddle point.
The one critical point for this function therefore corresponds to a saddle point, which occurs at the point (0, 0, f(0, 0)) = (0, 0, 3).
Self-critique (if necessary):
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Question: `q002. Consider f(x,y) = x^2 - y^2 + xy/16. Find the critical points, and classify each point as either a relative maximum, relative minimum, or saddle point.
Your solution:
Confidence rating:
Given Solution: f_x = 2 x + y / 16 and f_y = -2 y + x / 16.
Setting the two equal to zero and solving we find that (0, 0) is our only critical point.
f_xx = 2 and f_yy = -2. This indicates that the y = 0 'slice' of the graph has a minimum while the x = 0 'slice' has a maximum. The critical point therefore yields a saddle point.
We could evaluate f_xy (which is 1/16) and test for a saddle point, but we've already seen that we do have a saddle point. So we normally wouldn't bother. However, just to illustrate that the test for a saddle point works in this case, let's do the test.
f_xx * f_yy - f_xy ^ 2 = 2 * (-2) - (1/16)^2 = -4 - 1/16. This is clearly < 0, indicating what we already knew, that the point (0, 0) is a saddle point.
Note how f_xx * f_yy came out negative, since the signs of the two second derivatives differed. As soon as this happened the left-hand side was doomed to be negative, since the only other term - f_xy ^ 2 (being the negative of a square) cannot be positive.
Self-critique (if necessary):
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Question: `q003. Find the least squares regression line for the set of points {(4,-2), (3,-1), (0,0), (-1,3), (-2,1), (-3,2)}.
Your solution:
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Given Solution:
If y = a x + b, then the points on the line corresponding to x = 4, 3, 0, -1, -2, -3 have respective y coordinates 4 a + b, 3 a + b, b, -a + b, -2 a + b and -3 a + b. These differ from the given y coordinates by respective amounts 4 a + b - (-2), 3 a + b - (-1), b, -a + b - 3, -2 a + b - 1 and -3a + b - 2.
The sum of the squares of these differences is thus
(4 a + b - (-2))^2 + (3 a + b - (-1))^2 + b^2 + (-a + b - 3)^2 + ( -2 a + b - 1 )^2 + (-3a + b - 2))^2 = 39 a^2 + 2 a b + 44 a + 6 b^2 - 6 b + 19
In order to find the regression line we will find the values of a and b that minimize this sum.
For the sake of convenient notation let
f(a, b) = 39 a^2 + 2 a b + 44 a + 6 b^2 - 6 b + 19.
f_a = 78 a + 2 b + 44
f_b = 12 b + 2 a - 6
Solving simultaneously we obtain approximate values a = -.58 and b = .60.
We conclude that the best-fit equation, with a and b accurate to 2 significant figures, is
y = -.58 x + .60.
Self-critique (if necessary):
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Question: `q004. Consider these following functions, at each of which D = 0 at a critical point. Show whether each of the following is true or false:
f_xx is positive and f_yy is negative. The intersection of the graph of f(x, y) with the x-z plane is a curve with a relative min at (0, 0); the intersection with the x-y plane has a relative max at the same point.
Consider the line parameterized by x = t cos(theta), y = t sin(theta). Along this line we have z = t^4 cos^2(theta) sin^2(theta). For any value of theta except a multiple of 2 pi the function z vs. t has a relative minimum at t = 0.
Your solution:
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Given Solution:
Self-critique (if necessary):
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