Using Lagrange Multipliers to Solve Constrained Optimization Problems

[ElijahRockers]

Homework Statement

f(x,y) = y²-x², g(x,y) = x²/4 +y²=9

Homework Equations

$\nabla f = \lambda \nabla g$

$-2x = \lambda \frac{x}{2}$
$2y = 2\lambda y$
$\frac{1}{4} x^2 + y^2 = 9$

The Attempt at a Solution

I arrived at the three equations above. So according to the first equation, lambda can equal -4. According to the second equation, it can equal 1. After this, I am algebraically lost. The x's and y's cancel themselves out from the first two equations. What does this mean?

 

[Dick]

−2x=λx/2 means EITHER λ=(-4) OR x=0. You have to check both options.

 

[Ray Vickson]

Homework Statement

f(x,y) = y²-x², g(x,y) = x²/4 +y²=9

Homework Equations

$\nabla f = \lambda \nabla g$

$-2x = \lambda \frac{x}{2}$
$2y = 2\lambda y$
$\frac{1}{4} x^2 + y^2 = 9$

The Attempt at a Solution

I arrived at the three equations above. So according to the first equation, lambda can equal -4. According to the second equation, it can equal 1. After this, I am algebraically lost. The x's and y's cancel themselves out from the first two equations. What does this mean?

If x ± 0 then λ = -4, so in the second equation you must have y = 0.

RGV

 

[Dick]

If x ± 0 then λ = 4, so in the second equation you must have y = 0.

RGV

What are you talking about?

 

[Ray Vickson]

What are you talking about?

I had the typo λ = 4 instead of the correct λ = -4, but that still implies we need y = 0 to satisfy the second equation (which would be 2y = -8y).

RGV

 

[ElijahRockers]

Ok, well a lambda of -4 makes the other equation untrue. So lambda can not be -4 then, right?

 

[Dick]

Ok, well a lambda of -4 makes the other equation untrue. So lambda can not be -4 then, right?

Yes, it can. lambda can be -4 if y is zero.

 

[ElijahRockers]

Duh, ok, I think I got it. Thanks again.