course Mth 272 8/1 5 pm 030.
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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 - 4 fxy^2 = 2 * 20 - 4 * 6^2 = 8. 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): The given answer has a wrong Second Partials Test and wrong algebra (fxx * fyy - 4 fxy^2 = 2 * 20 - 4 * 6^2 = 8). My answer does not match the given answer, but since the given answer is mathematically incorrect, I believe my answer that I found with the book? Second Partials Test and information about d is correct.
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********************************************* 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. YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY Your solution: To find the extrema, we first need to find F?, F?x, F?y, F?, and F?y using our derivative rules. We find that F? = 3x^2 -6x +3, F?x = 6x -6, and F?y = 0. We also find that F? = 3y^2 + 12y +12, and F?y = 6y+12. Now we can set F? and F? equal to 0. So we find that 3x^2 -6x +3 = 0 and 3y^2 +12y +12 = 0. Solving these algebraically, we get the point(1,-2). Since F?y = 0, we find that the Second Partials Test tells us nothing. confidence rating: 3 ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^
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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 - 4 fxy^2 = 0, which is inconclusive--it tells us nothing about max, min or saddle point. &&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&& Self-critique (if necessary): ------------------------------------------------ Self-critique Rating: ********************************************* Question: `qAt what point(s) did the second-partials test fail? YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY Your solution: From the previous solution, the Second Partials test fails at the point(1,-2) because F?y = 0. confidence rating: 3 ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^
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Given Solution: `a 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 - 4 fxy^2 = 0, which is inconclusive--it tells us nothing about max, min or saddle point.