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

  • #1

Homework Statement


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

Homework Equations





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.

Does that sound about right?
 

Answers and Replies

  • #2
HallsofIvy
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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, [itex]\lambda= -1/4[/itex]. Similarly, if z is not 0, [itex]\lambda= 1[/itex]. If [itex]\lambda= 1[/itex], the first equation cannot be satisfied but if [itex]\lambda= -1/4[/itex] the first equation gives x= 4/5. Putting x= 4/5, z= 0 into [itex]x^2+ 4y^2+ z^2= 4[/itex] gives [itex]16/25+ 4y^2= 4[/itex] so [itex]y^2= 84/25[/itex] and then [itex]y= 2\sqrt{21}/5[/itex]. [itex](4/5, 2\sqrt{21}/5, 0)[/itex] also satisfies those equations. What is [itex]F(4/5, 2\sqrt{21}/5, 0)[/itex]?
 
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  • #3
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, [itex]\lambda= -1/4[/itex]. Similarly, if z is not 0, [itex]\lambda= 1[/itex]. If [itex]\lambda= 1[/itex], the first equation cannot be satisfied but if [itex]\lambda= -1/4[/itex] the first equation gives x= 4/5. Putting x= 4/5, z= 0 into [itex]x^2+ 4y^2+ z^2= 4[/itex] gives [itex]16/25+ 4y^2= 4[/itex] so [itex]y^2= 84/25[/itex] and then [itex]y= 2\sqrt{21}/5[/itex]. [itex](4/5, 2\sqrt{21}/5, 0)[/itex] also satisfies those equations. What is [itex]F(4/5, 2\sqrt{21}/5, 0)[/itex]?
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.

Sound about right?
 

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