Question Solved1 Answer Find the local maximum and minimum values and saddle point(s) of the function f(x, y) = 6xy2 – 2x3 – 3y4 Note: the second derivative test at (0, 0) fails. You can consider f(x, 0) = -2x), f(0, y) = -3y4 Select one: O A. (1, 1) is a local maximum, (1, -1) is a local maximum, (0, 0) is a local minimum. O B. (1, 1) is a local minimum, (1, -1) is a local maximum, (0, 0) is a saddle point. O C. (1, 1) is a local maximum, (1, -1) is a local maximum, (0, 0) is a saddle point. O D. (1, 1) is a local maximum, (1, -1) is a local maximum, (0, 0) is a local minimum. O E. (1, 1) is a local maximum, (1, -1) is a local minimum, (0, 0) is a local maximum. Find the points on the cone z? = 2x2 + y2 that are closest to the point (6,2,0). Select one: O A. (2, 1, 101/2) O B.(3, 1, +3) O C. (2, 1, +3) O D. (1, 2, +3) O E. no solutions Find the volume of the largest rectangular box in the first octant with three faces in the coordinate planes and one vertex in the plane x + 3y + 6z = 18. Round to two decimal places. Answer:

NGMA7Q The Asker · Calculus

Transcribed Image Text: Find the local maximum and minimum values and saddle point(s) of the function f(x, y) = 6xy2 – 2x3 – 3y4 Note: the second derivative test at (0, 0) fails. You can consider f(x, 0) = -2x), f(0, y) = -3y4 Select one: O A. (1, 1) is a local maximum, (1, -1) is a local maximum, (0, 0) is a local minimum. O B. (1, 1) is a local minimum, (1, -1) is a local maximum, (0, 0) is a saddle point. O C. (1, 1) is a local maximum, (1, -1) is a local maximum, (0, 0) is a saddle point. O D. (1, 1) is a local maximum, (1, -1) is a local maximum, (0, 0) is a local minimum. O E. (1, 1) is a local maximum, (1, -1) is a local minimum, (0, 0) is a local maximum. Find the points on the cone z? = 2x2 + y2 that are closest to the point (6,2,0). Select one: O A. (2, 1, 101/2) O B.(3, 1, +3) O C. (2, 1, +3) O D. (1, 2, +3) O E. no solutions Find the volume of the largest rectangular box in the first octant with three faces in the coordinate planes and one vertex in the plane x + 3y + 6z = 18. Round to two decimal places. Answer:
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Transcribed Image Text: Find the local maximum and minimum values and saddle point(s) of the function f(x, y) = 6xy2 – 2x3 – 3y4 Note: the second derivative test at (0, 0) fails. You can consider f(x, 0) = -2x), f(0, y) = -3y4 Select one: O A. (1, 1) is a local maximum, (1, -1) is a local maximum, (0, 0) is a local minimum. O B. (1, 1) is a local minimum, (1, -1) is a local maximum, (0, 0) is a saddle point. O C. (1, 1) is a local maximum, (1, -1) is a local maximum, (0, 0) is a saddle point. O D. (1, 1) is a local maximum, (1, -1) is a local maximum, (0, 0) is a local minimum. O E. (1, 1) is a local maximum, (1, -1) is a local minimum, (0, 0) is a local maximum. Find the points on the cone z? = 2x2 + y2 that are closest to the point (6,2,0). Select one: O A. (2, 1, 101/2) O B.(3, 1, +3) O C. (2, 1, +3) O D. (1, 2, +3) O E. no solutions Find the volume of the largest rectangular box in the first octant with three faces in the coordinate planes and one vertex in the plane x + 3y + 6z = 18. Round to two decimal places. Answer:
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[Q.1] f(x,y)=6xy^(2)-2x^(3)-3y^(4) If D > 0 and f_(xx)(a,b) < 0Then f(x,y) has a delativetheta=-12 x(12 x-36y^(2))-(12 y)^(2)Check at (1,-1)So thent it is a donbiful caseD=-12(12-36)-(12)^(2)quadHence f has neither a D=3x^(2)+2y^(2)-12 x-4y+40 maximum hor a minimuon value at the arigin.f(0,y)0-f(0,0)=-3y^(4) < 0,AA y" or "_(r)=6x-12=0=>x=2So f hoe a ... See the full answer