The plane x+y+2z=18 intersects the paraboloid z=x2+y2 in an ellipse. Find the points on this ellipse that are nearest to and farthest from the origin.

Question

SoosteethicU · Accepted Answer

It suffices to find the extreme values of the square of the distance to the origin D=x2+y2+z2, subject to the constraints g=x+y+2z=18andh=x2+y2−z=0. μ  By Lagrange Multipliers,▽D=λ▽g+μ▽h. ⇒&amp;lt;2x,2y,2z&amp;gt;=λ&amp;lt;1,1,2&amp;gt;+μ&amp;lt;2x,2y,−1&amp;gt;.  Equating like entries, 2x=λ+2μx 2y=λ+2μy 2z=2λ−μ  Since the last equation can be rewritten as μ=2λ−2z, we now have 2x=λ+2x(2λ−2z)=⇒2x=λ(4x+1)−4xz 2y=λ+2y(2λ−2z)=⇒2y=λ(4y+1)−4yz So,2x+4xz4x+1=λ=2y+4yz4y+1⇒2x1+2z4x+1=2y1+2z4y+1.   *If 1+2z=0≤⇒z=−12,then h=x2+y2=−12 has no solution.  So, we can ignore this case.  Now, we can freely divide both sides by (1+2z):  2x4x+1=2y4y+1 ⇒2x(4y+1)=2y(4x+1) ⇒x=y.  Substitute this into the constraint equations: x+x+2z=18⇒x+z=9 x2+x2−z=0⇒z=2x2  So,x+2x2=9. ⇒2x2+x−9=0 ⇒x=−1+734or−1−734  So, (

The plane x+y+2z=18 intersects the paraboloid z=x^2+y^2 in an ellipse.Find the points on this ellipse that are nearest to and farthest from the origin

Answered question

Answer & Explanation

New Questions in Calculus 2