# Find maximum on elipsoid using implicit function theorem…again

I feel like im drowning this site with question about implicit function theorem but I really do not understand how I can find the differential.

we are given elipsoid $x^2+y^2+z^2+xy+yz-54=0$

We are asked to show that the maximal value of $z$ on the elipsoid is at $(x,y,z)=(3,-6,9)$ using the implicit function theorem

if we define $F(x,y,z)=x^2+y^2+z^2+xy+yz-54$ we will see that the gradient (differential) is:

$(2x+y,2y+x+z,2z+y)$ meaning if $z \neq -\frac{y}{2}$ then we can represent $z=z(x,y)$ as a function of $x$ and $y$ according to implicit function theorem.

the maximal value of $z$ will be attained when $z'(x,y)=0$ or at a point where $z=-\frac{y}{2}$ (at which case, we cant represent $z$ as a function of $x$ and $y$.)

how do we find the differential of $z$?

I mean we want, $\frac{\partial z}{\partial x} = -\frac{F_x}{F_z} = \frac{-2x-y}{2z+y}=0$ and $\frac{\partial z}{\partial y} = -\frac{F_y}{F_z} = \frac{-2y-x}{2z+y}=0$

That would imply $x=0,y=0$ and $z=\sqrt{54}$. But perhaps there are higher values of $z$ where we cant express $z$ as a function...how do we find those?

• It's interesting that: $2x + y = 0$ , $2y + x + z = 0$, and $x^2 + y^2 + z^2 + xy + yz = 54$ has the solution set: $( 3,-6,9)$ , $(-3,6,-9)$ the maxima and minima in the problem. (Using wolfram alpha) – Alan Apr 6 '14 at 0:28
• The condition that the planes intersect might also make the dx and dy differential vanish. So, (t,-2t,3t) is the line which intersects the ellipsoid at t=3 and t = -3 , giving the coordinates ( 3,-6,9), (-3,6,-9) maxima and minima. – Alan Apr 6 '14 at 1:07
• for Max,$z=\dfrac{-y + \sqrt{216-4x^2-3y^2-4xy}}{2},\frac{\partial z}{\partial x} = 0,\frac{\partial z}{\partial y} = 0$,you will get the result at once. – chenbai Apr 6 '14 at 8:04
• Could you explain how you got to this result? – Oria Gruber Apr 6 '14 at 9:25

$(x,y,z)=(3,-6,9)$ is the correct answer. the $\frac{\partial z}{\partial y}$ written in the question is wrong.