Question

Use Lagrange multipliers to find the points on the cone 22 - x2 + y2 that are closest to (14, 10, 0). (x, y, z) - ) (smaller z-value) (larger z-value) DETAILS SESSCALC2 11.6.042. MY NOTES ASK YOUR TEAC At what point on the paraboloid y = x + 2? Is the tangent plane parallel to the plane 7x + 4y + 72 - 4? (If an answer does not exist, enter DNE.) (*. V. 2) - (L

Answer 1

ANSWER

* use lagrange multipliers to find the points on the cone z2 = x² + y 2 that are closest to (14, 10,0) Given cuve is z²= x² + y ² 6 suppose P(x,y,z) be the point on a which is nearer to (14110,0) Then op?= (-14)²+(4-10)²7 2² Considee timy, z)= op? (x-14)2 +Cy-10)42 and $(214,2)= x² + y²_2 Define F(x,y)=f+1$ = (x–14)+(7-10)+27) (x2+y? 27 Fx = 2(x - 14)+ X(2X) = 2 (y-10)+1(24) Fz=aF - 22+^(-32) Suppose Fx =o → 2CX-14-)+(2x)}=0->@x)) = -2(674) = A (x4) 2x -) = (x+4) =ƏR эх Fy= of 2 X Fy=O=> 269-10)+1(2y)=0 >> - Mzy)= 2(7-10) od = f(y-10) ->=9-10 dy Ę = 0 = 22 + x (-22)-0 > 16-22)- У. x=1

-dur-14 Y-10 y 1-14 y-10 y -1 1-14 -- -1 X-14=-x +2=14=> 2X= 14 x Y-10 y 1 >> Y-10=-y => 4 ty=10+2y=lo TY=10 z2-n2+y2 z²= 72+ 103 z?- 49+100 Z-149 25 + 12.20 Therefore (7, 10,-12) and (7, 10, 12) are the points on 0 Whose distance from (14,10,0) is minimum. At what point on the palaboloid y=x2+2? is the tangent plane parallel to the plane 74 +44 +72=4

Given paraboid y=xrz? 8 given plane 7x+4Y+72=4 paeaboid. y=x²7 22 fCm, 4,2 )= x2+ zz-y. find normal vector to pauuboid noomal vector - DFLXlYız) = 249-3 +224 x²0-y+22 = Qui_jtazu qui-jtazk Normal vector to the given plane fx+4y +72=4 is $i+4; +7 since, we want the tangent plane at the to be parallel to plane x+y+z=1 then the normal vector of (ury, z)= 2x1 ; +azů has to be paeallel with the normal rector 71 +4;+75 That means anitatzk must be constant multiple of 7i+45+7K A ani]+22F=c(71 +43 +75) Gompale cofficients. -1=46 Cecele 6=-1 point 4

O ) Q~ 개그 C 2257 오조스크 Z= 8 plugging into the fan of the surface = 2 8-1)+()

5) 9 = 44+ 49 64 98 64 19: CEG 3 2 po int (Miyiz) - 19. ) 꾸 8