Solutions to Practice Test 3

60z 402001y 23 543x21We first find the critical points by setting the first partials equal to 0.f x (x, y) =16xy − 2xy 2 − 3x 2 y = xy(−2y + 16 − 3x) = 0 ® x = 0ory = 0or−2y + 16 − 3x = 0f y (x, y) =8x 2 − 2x 2 y − x 3 = −x 2 (−8 + 2y + x) ® x = 0or−8 + 2y + x = 0x = 0 and y = 0 are not in the domain under consideration, so we ignore these values.−2y + 16 − 3x = 0 and −8 + 2y + x = 0 ® x = 4 and y = 2, and we have the critical point (4, 2).f(4,2) =64We now find the value of the function along the edges of D.L 1 :={(x,y)|1 ≤ x ≤ 5, y = 1} ® f(x,1)=7x 2 − x 3 = −x 2 (−7 + x).We check for the extreme values of this function by taking the derivative:f (x,1)=14x − 3x 2 = −x(−14 + 3x) = 0 ® x = 14 U 31372,1)= = 50.81527f( 143The function must be evaluated at the endpoins x = 1 and x = 5.f(1,1) =6f(5,1) =50L 2 :={(x,y)|x = 5,1 ≤ y ≤ 3} ® f(5,y) =75y − 25y 2 = −25y(−3 + y)We check for the extreme values of this function by taking the derivative:f (5, y) =75 − 50y ® y = 3 U 2f(5, 3 225)= = 56.252 4The function must be evaluated at the endpoins y = 1 and y = 3.f(5,1) =50f(5,3) =0L 3 :={(x,y)|1 ≤ x ≤ 5, y = 3} ® f(x,3)=15x 2 − 3x 3 = −3x 2 (−5 + x)We check for the extreme values of this function by taking the derivative:f (x,3)=30x − 9x 2 ® x = 10 U 3f( 103,3)=5009= 55.556The function must be evaluated at the endpoins x = 1 and x = 5.f(1,3) =12f(5,3) =0L 4 :={(x,y)|x = 1,1 ≤ y ≤ 3} ® f(1,y) =7y − y 2 = −y(−7 + y)We check for the extreme values of this function by taking the derivative:f (1, y) =7 − 2y ® y = 7 U 2f(1, 7 49)= = 12.252 4The function must be evaluated at the endpoins y = 1 and y = 3.f(1,1) =6f(1,3) =12The absolute minimum is f(5,3) =0, and the absolute maximum is f(4,2) =64.