Need help with question 2. laxima, and no saddlepoints2. Find the point on the graph of z=-x^2-^2-xy that is the farthest above the plane 5x+4 y + -3 (use verticaldistance, not overall distance). How far above the plane is that point?Select one:C.a. 5C.b. 10C.с. 12d. 3C.e. 73. Find the local extrema of f (x, y) =2xy+ 1/2x 1/y (Select all that apply.)Select one or more:a. local maximum at (-1/2,1)b. local maximum at (-1,1/4)c. local minimum at (0,0)d. saddle point at (1,1/4)e. local minimum at (-1/2,1)4.Find the minimum value of the integral below, where (x,y) lies on the unit circle.S.2z dzSelect one:L.L L

Name: Need help with question 2. laxima, and no saddlepoints2. Find the poin
Uploaded: 2024-03-20T07:07:37.000Z
Channel: Bartleby
Description: Need help with question 2. laxima, and no saddlepoints2. Find the point on the graph of z=-x^2-^2-xy that is the farthest above the plane 5x+4 y + -3 (use verticaldistance, not overall distance). How far above the plane is that point?Select one:C.a.

Given Equation of graph is z=-x2-y2-xy (1) and equation of plane is 5x+4y+z=-3⇒5x+4y+z+3=0 (2)We know that, Distance from the point (x, y, z) to a plane with equation (2) is D(x, y, z)=5x+4y+z+33 (3) Since point lies on the graph, so distance is the function of x and y. Hence, by putting the value of z from equation (1) to equation (2) we have f(x, y)=5x+4y-x2-y2-xy+33 (4) Since the graph lies above the plane then f(x, y) is positive. hence, f(x, y)=5x+4y-x2-y2-xy+33 to find the point of maximum we have to partially differentiate the function f(x, y) and equating them to zero. So we have fx(x, y)=0 ⇒5-2x-y=0 (5) fy(x, y)=0⇒4-2y-x=0 (6) From equation (5) and (6), we have x=2,y=1 putting the value of x and y in equation (1) z=-7 Hence the point on the graph is (2, 1, -7) and maximum distance is D(2, 1, -7)=5×2+4×1-7+33D(2, 1, -7)=3.33 Answer: The point on the graph is (2, 1, -7) Distance between plane and point is 3.33. Hence option (d) is correct.