Question

The method of Lagrange multipliers assumes that the extreme values exist, but that is not always the case. Show that the problem of finding the minimum value of f(x, y) = x^2 + y^2 subject to the constraint can be solved using Lagrange multipliers, but does not have a maximum value with that constraint.

Answer 1

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Explanation:

Given the function,

F(x, y)=x²+y²

The La Grange is theorem

Solve the following system of equations.

∇f(x, y)= λ∇g(x, y)

g(x, y)=k

Fx=λgx

Fy=λgy

Fz=λgz

Plug in all solutions, (x,y), from the first step into f(x, y) and identify the minimum and maximum values, provided they exist and

∇g≠0 at the point.

The constant, λ, is called the Lagrange Multiplier.

F(x, y)=x²+y²

∇f= 2x i + 2y j

So, given the constraint is xy=1.

g(x, y)= xy-1=0

∇g= y i + x j

gx= y. And gy=x

So, here is the system of equations that we need to solve.

Fx=λgx; 2x=λy. Equation 1

Fy=λgy; 2y=λx. Equation 2

xy=1

Solving this

x=λy/2. From equation 1, now substitute this into equation 2

2y=λ(λy/2)

2y=λ²y/2

2y-λ²y/2 =0

y(2-λ²/2)=0

Then, y=0. Or (2-λ²/2)=0

-λ²/2=-2

λ²=4

Then, λ= ±2

So substitute λ=±2 into equation 2

2y=2x

Then, y=x

So inserting this into the constraint g will give

xy=1. Since y=x

x²=1

Therefore,

x=√1

x=±1

Also y=x

Then, y=±1

Therefore, there are four points that solve the system above.

(1,1) (-1,-1) (1,-1) and (-1,1)

The first two points (1,1) (-1,-1) shows the minimum points because they show xy=1

The other points does not give xy=1

They give xy=-1.

Now,

F(x, y)=x²+y²

F(1,1)=1²+1²

F(1,1)=2

F(-1,-1)= (-1) ²+(-1)²

F(-1,-1)=1+1

F(-1,-1)=2

Then

F(1,1)= F(-1,-1)=2 is the minimum point

This gives the same four points as we found using Lagrange multipliers above.