Use lagrange multipliers to find the max and min values of the func f(x,y)=xy subject to the constraint 1=(x^2)/8 + (y^2)/2
I know how to set up..
i got y=Lx/4 and x=Ly
now im lost!!!
Please help with detail because im an idiot.
From the second equation, find
L=-x/y, substitute into the first to get
y-x^2/4y=0
which gives x=2y
Substitute in the third equation to get
4y^2/8+y^2/2-1=0
which gives y=1 which => x=2.
So (2,1) is a maximum/minimum.
Since the constraints have x^2 and y^2,
so
(-2,1), (-2,-1) and (2,-1) are all feasible solutions.
Locate the maxima and minima by substitution into F(x,y).
why is L=-x/y why is this negative??
L can be any number, positive, negative or even zero. It is present to ensure that the constraints are satisfied.
For example, we could have equally written the constraint equation as
L(1-x^2/8-y^2/2), then L would have been positive.
Don't worry, it's completely normal to get stuck on concepts like Lagrange multipliers. I'm here to help you understand the process step by step.
1. Start by setting up the Lagrangian function. The Lagrangian is defined as the function f(x, y) minus the product of the constraint equation and the Lagrange multiplier λ (lambda). In this case, the Lagrangian function is:
L(x, y, λ) = f(x, y) - λ(g(x, y))
where f(x, y) = xy is the objective function, g(x, y) = 1 - (x^2)/8 - (y^2)/2 is the constraint equation, and λ is the Lagrange multiplier.
2. The next step is to find the partial derivatives of the Lagrangian with respect to x, y, and λ:
dL/dx = y - λ(dx/dx)
dL/dy = x - λ(dy/dx)
dL/dλ = g(x, y)
Since dx/dx is simply 1 and dy/dx is 0 (because y is not a function of x), the partial derivatives simplify to:
dL/dx = y - λ
dL/dy = x
dL/dλ = g(x, y)
3. To find the critical points, set all three partial derivatives equal to zero and solve the resulting system of equations. Start by setting dL/dx = 0 and dL/dy = 0:
y - λ = 0
x = 0
Using the constraint equation, g(x, y) = 1 - (x^2)/8 - (y^2)/2, we substitute x = 0:
1 - (0^2)/8 - (y^2)/2 = 0
1 - 0 - (y^2)/2 = 0
1 - (y^2)/2 = 0
4. Solve the equation from step 3 for y:
1 - (y^2)/2 = 0
(y^2)/2 = 1
y^2 = 2
y = ±√2
5. Substitute the values of x and y back into the constraint equation to find the corresponding values of λ:
For y = √2:
1 - (x^2)/8 - (√2^2)/2 = 0
1 - (x^2)/8 - 1 = 0
-(x^2)/8 = 0
x^2 = 0
x = 0
For y = -√2:
1 - (x^2)/8 - (-√2^2)/2 = 0
1 - (x^2)/8 - 1 = 0
-(x^2)/8 = 0
x^2 = 0
x = 0
6. Finally, substitute the values of x and y into the objective function to find the maximum and minimum values of f(x, y):
For x = 0 and y = √2:
f(x, y) = xy
f(0, √2) = 0 * √2 = 0
For x = 0 and y = -√2:
f(x, y) = xy
f(0, -√2) = 0 * -√2 = 0
Therefore, according to the Lagrange multiplier method, the maximum and minimum values of f(x, y) = xy under the constraint g(x, y) = 1 - (x^2)/8 - (y^2)/2 are both 0.
I hope this explanation helps you understand the process. Let me know if you have any further questions!