#$&* course Mth 277 12/6/2011 525pm If your solution to stated problem does not match the given solution, you should self-critique per instructions at
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Given Solution: f_x = 6 x - 5 y and f_y = 5 x + 2 y. Our critical point therefore occurs when 6x - 5y = 0 and 5x + 2y = 0. Solving simultaneously we get x = 0, y = 0 so our critical point is (0, 0). The second derivatives are f_xx = 6, f_yy = 5 and f_xy = -5. f_xx and f_yy are both positive, indicating that if the critical point is not a saddle point, it is a minimum f_xx * f_yy - f_xy^2 = 6 * 5 - (-5)^2 = 5, which is > 0 and indicates that the critical point is not a saddle point. The one critical point for this function therefore corresponds to a relative minimum, which occurs at the point (0, 0, f(0, 0)) = (0, 0, 3). &&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&& Self-critique (if necessary): OK ------------------------------------------------ Self-critique rating: ********************************************* Question: `q002. Consider f(x,y) = x^2 - y^2 + xy/16. Find the critical points, and classify each point as either a relative maximum, relative minimum, or saddle point. YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY Your solution: f_x = 2x + y/16 = 0 x = -y/8 f_y = -2y + x/16 0 = -2y + (-y/8)/16 y = 0 x = 0 f_xx = 2 f_yy = -2 f_xy = 1/16 f_xx + f_yy + f_xy = 2 - 2 - (1/16)^2 < 0 therefore we have a saddle point. confidence rating #$&*: ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^
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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): OK ------------------------------------------------ Self-critique rating: ********************************************* 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)}. YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY Your solution: I set up a table x,y,xy,x^2 4,2,8,16 3,-1,-3,9 0,0,0,0 -1,3,-3,1 -2,1,-2,4 -3,2,-6,9 1,7,-6,39 The first line is the variables to be found for each point The second line is the first point (4,-2) solved for each of the variables in the first line The third, 4th, 5th, 6th and 7th lines are the set of point (3,-1), (0,0), (-1,3), (-2,1), (-3,2), solved for the variables in the first line the 8th line is the sum of each of the each column I used the least squares regression line: a = [n*sum(xy) - sum(x)*sum(y)] / [n*sumx^2 - (sumx)^2] = -43/233 =app -0.185 is the x value b = (1/n)*[sumy - a*sumx] =appr 1.136 is the y value least squares regression y = -0.185x + 1.136 confidence rating #$&*: ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^
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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): hmm.. didn't do it the same way, but I stand by my answer ------------------------------------------------ Self-critique rating:3
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Given Solution: &&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&& Self-critique (if necessary): ------------------------------------------------ Self-critique rating:
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