Answer:
Incomplete question check attachment for complete question
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.
