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Section 26 HW Answers There is a different amt of work shown for different problems, the most is shown for the first 6 1) • fx (x, y) = −4x • fy (x, y) = −6y • Local MAX at (0, 0, f (0, 0)) equation (0, 0) fxx = −4 -4 fyy = −6 -6 fxy =0 0 2 D fxx × fyy − fxy 24 Since D is positive and fxx is Negative, (0, 0, f (0, 0)) is a local MAX 2) • fx (x, y) = 3x2 • fy (x, y) = 6y • Set fx (x, y) = 0 and fy (x, y) = 0 and solve the system of equations. 3x2 = 0 x2 = 0 x=0 6y = 0 Each eqn only has one variable, so we can solve them separately y=0 So we put that info together and know that the only pairs of numbers x, y that make both equations true at the same time is when x = 0 and y = 0. So our only critical point is (0, 0, f (0, 0)) • Critical Points: (0, 0, 0) • Test gives us no info about at (0, 0, f (0, 0)) equation (0, 0) fxx = 6x 0 fyy = −6 -6 fxy =0 0 2 D fxx × fyy − fxy 0 Since D = 0 this test gives no information 1 3) • fx (x, y) = 2x − 2y + 1 • fy (x, y) = −2x + 4y • Set fx (x, y) = 0 and fy (x, y) = 0 and solve the system of equations. 2x − 2y + 1 = 0 −2x + 4y = 0 4y = 2x 2y = x 2(2y) − 2y + 1 = 0 4y − 2y + 1 = 0 2y = −1 y = −1 2 Both Eqns are messy, start with one We have one variable in terms of the other :) Now that we know about y, let’s find the x value(s) 2( −1 2 ) =x −1 = x Add (−1, −1 , f (−1, −1 )) 2 2 to our CP list −1 • Critical Points: (−1, −1 2 , 2 ) −1 • Local Min at (−1, −1 2 , f (−1, 2 )) fxx fyy fxy D equation =2 =4 = −2 fxx × fyy − fxy 2 (−1, −1 2 ) 2 4 -2 4 −1 Since D is positive and fxx is positive there is a local MIN at (−1, −1 2 , f (−1, 2 )) 2 4) • fx (x, y) = 2x − y 2 (2x) • fy (x, y) = 2y − x2 2y • Set fx (x, y) = 0 and fy (x, y) = 0 and solve the system of equations. 2x − y 2 (2x) = 0 2y − x2 2y = 0 2(y − x2 y) = 0 2y(1 − x2 ) = 0 2y = 0 or 1 − x2 = 0 y = 0 or x = 1 or x = −1 When y = 0 2x − (0)2 (2x) = 0 2x = 0 x=0 When x = 1 2(1) − y 2 (2 · 1) = 0 2 = 2y 2 1 = y2 1 = y or −1 = y When x = −1 2(−1) − y 2 (2 · −1) = 0 y 2 (2) = 2 y2 = 1 y = 1 or y = −1 Both Equations are messy, start with either one: Plug each of these 3 possibilities back into the other eqn Add (0, 0, f (0, 0)) to the CP list Add (1, 1, f (1, 1)) and (1, −1, f (1, −1)) to the CP list Add (−1, 1, f (−1, 1)) and (−1, −1, f (−1, −1)) to the CP lis • Critical Points: (0, 0, −4) , (1, 1, −3) , (1, −1, −3), (−1, 1, −3) and (−1, −1, −3) • Local Min AT (0, 0) and Saddle Points AT (1, 1), (1, −1), (−1, 1), and (−1, −1) equation (0, 0) (1, 1) (1, −1) (−1, 1) (−1, −1) 2 fxx = 2 − 2y 2 0 0 0 0 fyy = 2 − 2x2 2 0 0 0 0 fxy = −4xy 0 -4 4 4 -4 D fxx × fyy − fxy 2 4 -16 -16 -16 -16 AT (0, 0) Since D is positive and fxx is positive there is a local MIN at (0, 0) ( all others, D negative implies Saddle point. ) 3 5) • fx = y − • fy = x − 4 x2 2 y2 • Critical Value (2, 1, f (2, 1)) • That critical value is actually a local Min: equation (2, 1) 8 fxx = 3 1 x 4 fyy = 3 4 y fxy =1 1 2 D fxx × fyy − fxy 3 Since D is positive and fxx is positive, there is a MIN at (2, 1) 6) • fx (x, y) = 2(x − 1) • fy (x, y) = 3y 2 − 2y − 1 • Set fx (x, y) = 0 and fy (x, y) = 0 and solve the system of equations. 2(x − 1) = 0 2(x − 1) = 0 x-1 = 0 x=1 3y 2 − 2y − 1 = 0 Each eqn only has one variable, so we can solve them separately 3y 2 − 2y − 1 = 0 (3y + 1)(y − 1) = 0 y = −1 3 or y = 1 −1 Now we put together all possible pairs. So our critical points are (1, −1 3 , f (1, 3 )) and (1, 1, f (1, 1)) −1 • Critical Points: (1, −1 3 , f (1, 3 )) and (1, 1, 0) −1 • Saddle Point: (1, −1 3 , f (1, 3 )) and Local MIN: (1, 1, f (1, 1)) fxx fyy fxy D equation =2 = 6y − 2 =0 fxx × fyy − fxy 2 (1, −1 3 ) 2 -4 0 -8 (1, 1) 2 4 0 8 4 8) • fx (x, y) = 4x3 − • fy (x, y) = 3y 2 − 3 yx2 3 xy 2 • Set the second eqn = 0 and we get x = q q 20/19 19 4 3 20/19 19 4 , , , f ( • CP ( 34 3 4 3 )) 1 y4 (yes, this is a mess) 10) • Notice this function has a restricted natural domain (so the cancelation is ok) • fx (x, y) = (x − 8) · 1 (2xy) x2 y • fy (x, y) = (x − 8) x12 y (x2 ) = + ln(x2 y) = 2(x−8) x + ln(x2 y) (x−8) y • Setting the second eqn = 0 implies x = 8 + ln(x2 y) = 0 implies 0 + ln(82 y) = 0. Then, plugging x = 8 in to 2(x−8) x 1 Therefore 1 = 82 y, so y = 64 1 1 • CP = (8, 64 , f (8, 64 )) 5