Question

Find the point on the graph of z = -22 - y2 - ty that is the farthest above the plane 5x + 4y + z = -3 (use vertical distance, not overall distance). How far above the plane is that point? Select one: a. 12 b. 5 C. 3 d. 10 e. 7

PLEASE ANSWER ALL PARTS AND SHOW WORK. THANK YOU!

Answer 1

solution : any point on 3D space is (0,9, 2) distance of Ca. 9, 2) from farny + Z=-3 is Do da Suruy+2+3 √52+42+1 dc & Y,Z) = Suruya Z+3 √42 We have to maximise a subject to constrains, g (n.), 2) x+y2 + xy + z = 0 By lagrange's much puers, 5 A (20+4) va 1 (22+) a = 'Juz "Wh2 = A(zyra) Yuuz 0 So, if a zary 25 => 3n = 6 => R=2 and yel ay thay D. There is only one point that is at a fosthest distance from the plone. putting the value of n=2 & y=1 in gen,y,z)=0 we ges za - 2²-12-(2) (1) -4-1-2=-7 So, the point is (2, 1,-2) and d(2,1,-3) = 5 (2) +4 (1)-7+3 ru 142 30, the answer is 10 10