Question

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1. The plane 4x – 3y +8z = 5 intersects the cone z? = x² + y2 in an ellipse. Find the points on this ellipse the highest and the lowest points (i.e. largest and smallest values of z) on the ellipse.

Answer 1

- Чи -3y + 8 = S , zex? у 2. We have to optimise - f(x,y,z) = z. 8( xy,z) = (х -3y +8 z-s Methad used 4 (x, y, z) = x2+y2-22. Lagrange Multipliers Now let - Р (ҳy,z) . f (x,y,z) + + 4 (x,y,z) + 4 ( ҡ, y, z) = 2 + 1 ( Чx – 3y+8z) + ( x+y2-22) ƏF = чa + 2 их чА + 2x = 0 —сі) 2 – 32 + 2 y =) - 3 A + 2 y =0 — Е 21 + 84 - 2kz |+84 - 2 2 =0 (б). (fer ки у ось – но, ар - 1 0 - Елом гох (і) Х - Блач ад « (11) — У 2 3 .

from equ (1) - 24 M Now plage these values in equ of plane 434-28) - 3.1. +4 489) = 5 -89 - 94 +4+32 = 54 397 2 ty=54 ) 39 7+8=1ou in Now plage values of my, z in equ of cone - Cate = 1-20)*+ (panj 14 Guatemal wieder Itt + 16 ) y u2 - cle 2 3) 1+6492 +167= 259" => 1=-, -ić

- for 8 = -1/5 from equliv? 39+(-1/2) +8=10u = M = -1 / 2 for d = 1 1 1 3 from equlin) - 39% (-) +8 = 10u =) dezza. Son (9,6 —— (-,-Ź) (2,4) –> (*1). Since - z = 1+ 22 h 1781 2u So for (a,m) -1 6-, - A) z = 1+81-42 28/01 - 573. - Max for (Au) - ż). z = 1+8xl vs 2x → Min