# Finding Max/Min Values on Functions of 3 Variables, Bounded by Ellipsoids

1. Nov 12, 2011

### TranscendArcu

1. The problem statement, all variables and given/known data
Find the absolute minimum and maximum of F(x,y,z) = x2 - 2x - y2 + z2 on the ellipsoid G(x,y,z) = x2 + 4y2 + z2 = 4

2. Relevant equations

3. The attempt at a solution
I was thinking of trying to solve this by using Lagrange multipliers. So, finding the gradients:

Fx = 2x - 2 = Gx = λ 2x
Fy = - 2y = Gy = λ 8y
Fz = 2z = Gz = λ 2z

From the first partial derivative I have 2x - 2 - λ2x = 0, which suggests x = 1/(1-λ). From the second partial derivative I have y(-2 - λ * 8) = 0, which suggests y = 0. Similarly, from the third partial derivative I have z(2 - λ * 2) = 0, which suggests z = 0. From G(1/(1-λ),0,0) I get λ = ((-/+) 1/2) + 1, or λ = 1/2 or 3/2.

Therefore, x = -2 or 2. Evaluating F(-2,0,0) = 8 and F(2,0,0) = 0. So, (-2,0,0) is our max and (2,0,0) is our min.

2. Nov 12, 2011

### HallsofIvy

Staff Emeritus
The two equations may suggest a variable is 0 but it doesn't follow that it must be! However, you can argue that if y is not 0, then, dividing both sides of the second equation by y, $\lambda= -1/4$. Similarly, if z is not 0, $\lambda= 1$. If $\lambda= 1$, the first equation cannot be satisfied but if $\lambda= -1/4$ the first equation gives x= 4/5. Putting x= 4/5, z= 0 into $x^2+ 4y^2+ z^2= 4$ gives $16/25+ 4y^2= 4$ so $y^2= 84/25$ and then $y= 2\sqrt{21}/5$. $(4/5, 2\sqrt{21}/5, 0)$ also satisfies those equations. What is $F(4/5, 2\sqrt{21}/5, 0)$?

Last edited: Nov 12, 2011
3. Nov 13, 2011

### TranscendArcu

Do you mean y = ± √(21)/5, because 16/25 + 4(2sqrt(21)/5)^2 ≠ 4. Anyway, presuming this is what you meant, I have to test the points (2,0,0), (-2,0,0), (4/5,sqrt(21)/5,0), (4/5,-sqrt(21)/5,0).

F(2,0,0) = 0
F(-2,0,0) = 8
F(4/5,sqrt(21)/5,0) = -9/5
F(4/5,-sqrt(21)/5,0) = -9/5

Therefore, my max is at (-2,0,0) and is 8. My minimums are at (4/5,sqrt(21)/5,0) and (4/5,-sqrt(21)/5,0) and are -9/5.