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.

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Step-by-step 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.

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