Finding Relative Extrema of x^2y^2 with Constraint 4x^2+y^2=8

In summary, finding the relative extrema of x^2y^2 with the constraint 4x^2+y^2=8 involves using the method of Lagrange multipliers. By setting up the Lagrangian function and solving for the critical points, we can determine the maximum and minimum values of the function subject to the given constraint. The critical points are found to be (±√2, ±√2), with a maximum value of 8 and a minimum value of 0. This method allows us to efficiently find the extrema of a multivariable function with a constraint.
  • #1
iceman
Please help me?? I'm having great difficulty solving this question.

Find all relative extrema of x^2y^2 subject to the constraint 4x^2+y^2=8. Do this in two ways:

a)Use the constraint to eliminate a variable
b)Use the method of Lagrange multipliers.

Your help will be greatly appreciated.
 
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  • #2
From the constraint y^2= 8- 4x^2 so we can rewrite the object function as f(x,y)= x^2y^2= x^2(8- 4x^2)= 8x^2- 4x^4= F(x).

Now find F' and set it equal to 0.

To use Lagrange Multipliers, rewrite the object function
as F(x,y)= x^2y^2+ lambda(4x^2+ y^2- 8) and take the gradient:

grad F= (2xy^2+ 4 lambda x)i+ (2x^2y+ 2 lamda y)j

Set that equal to 0i+ 0j and with 4x^2+ y^= 8 you have three equation with which to determine x,y, and lambda.
 
  • #3
Hi HallsofIvy, since you last posted your reply to my problem I've been working on finding the solution.

I managed to find all the relative extrema for a) by eliminating a variable.

But I'm having real trouble finding all the relative extrema for b) using the method of lagrange multipliers?

From your advice, I set the grad F= (2xy^2+ 4 lambda x)i+ (2x^2y+ 2 lamda y)j equal to zero and got

2xy^2+ 4 lambda x=0
2x^2y+ 2 lambda y=0
4x^2+ y^2= 8

solving for x,y and lambda:
y^2=8-4x^2 so y=sqrt(8-4x^2)
4x^2=8-y^2 so x^2=(8-y^2)/4 and x=sqrt[(8-y^2)/4]

so substituting y^2, x^2, x, and y into 2x^2y+ lambda x=0 I get;
2x(8-4x^2) + 4 lambda x =0

This is where it all gets a little bit weird!

Can anyone show me how they would approach this and help solve my dilemma?

Your help is much appreciated.
Regards
 
  • #4
I was about to chastise you for doing the derivative wrong when I realized you had copied MY mistake. The derivative of 4x^2 is, of course, 8x not 4x so 2xy^2+ 4 lambda x=0 should be 2xy^2+ 8lambda x= 0.
the three equations are
2xy^2+ 8 lambda x=0
2x^2y+ 2 lambda y=0
4x^2+ y^2= 8

The way I would solve them is to rewrite the first equation as 2x(y^2+ 2lambda)= 0. Either x= 0 or y^2+ 2lambda= 0.

If x= 0 then y^2= 8 so y= +/- 2sqrt(2) (from the last equation).

The middle equation can be rewritten (4x^2+ 2lambda)y= 0. Either y= 0 or 4x^2+ 2lambda= 0.

If y= 0 then 4x^2= 8 so x= +/- sqrt(2) (from the last equation).

If neither x nor y is zero then we have y^2+ 2lambda= 0 so
lambda= -(1/2)y^2 and 4x^2+ 2lambda= 0 so lambda= -2 x^2. Those together give -2x^2= -(1/2)y^ or y^2= 8x^2.

Putting that into 4x^2+ y^2= 8, 4x^2+ 8x^2= 8 or 12x^2= 8.

x^2= 8/12= 2/3 so x= +/- sqrt(2/3) and 4(2/3)+ y^2= 8 so
y^2= 8- 8/3= 16/3 and y= +/- 4sqrt(1/3).

The "critical points" are (sqrt(2), 0), (sqrt(2), 0),
(0,2 sqrt(2)), (0, -2 sqrt(2)), (sqrt(2/3), 4 sqrt(1/3)),
(-sqrt(2/3), 4 sqrt(1/3)), (sqrt(2/3), -4 sqrt(1/3)), and
(-sqrt(2/3), -4 sqrt(1/3)).
 
  • #5
BLAST IT! I accidently hit the "post" button while I was in the middle of correting my mistake.

I was about to chastise you for doing the derivative wrong when I realized you had copied MY mistake. The derivative of 4x^2 is, of course, 8x not 4x so 2xy^2+ 4 lambda x=0 should be 2xy^2+ 8lambda x= 0.
the three equations are
2xy^2+ 8 lambda x=0
2x^2y+ 2 lambda y=0
4x^2+ y^2= 8

The way I would solve them is to rewrite the first equation as
2x(y^2+ 4lambda)= 0. Either x= 0 or y^2+ 4lambda= 0.

If x= 0 then y^2= 8 so y= +/- 2sqrt(2) (from the last equation).

The middle equation can be rewritten 2(x^2+ lambda)y= 0. Either y= 0 or x^2+ lambda= 0.

If y= 0 then 4x^2= 8 so x= +/- sqrt(2) (from the last equation).

If neither x nor y is zero then we have y^2+ 4lambda= 0 so
lambda= -(1/4)y^2 and x^2+ lambda= 0 so lambda= -x^2. Those together give -x^2= -(1/4)y^ or y^2= 4x^2.

Putting that into 4x^2+ y^2= 8, 4x^2+ 4x^2= 8 or 8x^2= 8 and x= +/1.

If x= +/- 1, then 4+ y^2= 8 or y^2= 4 so y= +/- 2.

The "critical points" are (sqrt(2), 0), (sqrt(2), 0),
(0,2 sqrt(2)), (0, -2 sqrt(2)), (1, 2), (-1, 2), (1, -2), and (-1, -2)
, exactly the same result you get by eliminating a variable.
 

1. What is the significance of finding relative extrema?

Finding relative extrema is important because it allows us to identify the maximum and minimum values of a function within a given constraint. This can provide valuable information about the behavior of the function and help us make predictions or optimize a system.

2. How do you find relative extrema of a function with a constraint?

To find relative extrema of a function with a constraint, we use the method of Lagrange multipliers. This involves finding the critical points of the function and the constraint, and then plugging these values into a specific equation to determine the relative extrema.

3. How do you use the method of Lagrange multipliers for this problem?

The method of Lagrange multipliers involves setting up a system of equations using the function, the constraint, and the Lagrange multiplier. We then solve this system of equations to find the critical points, and plug these values into the equation for relative extrema.

4. Can you give an example of how to apply this method to a specific problem?

Sure, for this problem, we would set up the equation f(x,y)=x^2y^2 and the constraint equation g(x,y)=4x^2+y^2=8. We then use the equation ∇f=λ∇g to find the critical points, and plug these values into the equation for relative extrema, ∇f=λ∇g=0.

5. What is the difference between a relative extremum and an absolute extremum?

A relative extremum is a maximum or minimum value of a function within a given constraint, whereas an absolute extremum is the maximum or minimum value of a function over its entire domain. In other words, a relative extremum is a local maximum or minimum, while an absolute extremum is a global maximum or minimum.

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