Question (1 point) The plane x +y + 2z = 8 intersects the paraboloid z = x2 + y2 in an ellipse. Find the points on this ellipse that are nearest to and farthest from the origin. Point farthest away occurs at ( Point nearest occurs at

OUVFLX The Asker · Calculus

Transcribed Image Text: (1 point) The plane x +y + 2z = 8 intersects the paraboloid z = x2 + y2 in an ellipse. Find the points on this ellipse that are nearest to and farthest from the origin. Point farthest away occurs at ( Point nearest occurs at

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BOLDSL

Let (x,y,z) be any point on the ellipse.Then the distance from (x,y,z) to the origin is given by d=sqrt((x-0)^(2)+(y-0)^(2)+(z-0)^(2))=sqrt(x^(2)+y^(2)+z^(2))For simplicity take the objective function as f(x,y,z)=d^(2)=x^(2)+y^(2)+z^(2), subject to the constraints g(x,y,z)=x+y+2z=8 and h(x,y,z)=x^(2)+y^(2)-z.Now from Lagrange's Multiplier Method{:[grad f=lambda grad g+mu grad h],[Longrightarrow(2x","2y","2z)=lambda(1","1","2)+mu(2x","2y","-1)],[Longrightarrow(2x","2y","2z)=(lambda+2x mu","lambda+2y mu","2lambda-mu)]:}Compairing both sidesSubstracting (ii) from (i)){:[2(x-y)=2mu(x-y)],[Longrightarrow2(x-y)(mu-1)=0]:}Now if mu=1, then lambda=0Longrightarrow z=-(1)/(2)Then x^(2)+y^(2)=-(1)/(2) ... See the full answer