min 2x+y subject to: x+y+z=1 and y^2+z^2=4
Any help would greatly be apprecaited.
y^2+z^2=4 --->
put y = 2 cos(theta) and
z = 2 sin(theta)
x+y+z=1 ---->
x = 1-2(cos(theta) + sin(theta))
2x + y =
2 - 2 cos(theta) - 4 sin(theta)
It's not difficult to find the minimum of this function!
You can also use lagrange multipliers...
To find the minimum value of the function 2x + y, subject to the constraints x + y + z = 1 and y^2 + z^2 = 4, you can use the method of Lagrange multipliers.
1. Start by writing the Lagrangian function L(x, y, z, λ1, λ2) as follows:
L(x, y, z, λ1, λ2) = 2x + y + λ1(x + y + z - 1) + λ2(y^2 + z^2 - 4)
2. Take partial derivatives of L with respect to x, y, z, λ1, and λ2 and set them equal to zero to find the critical points:
∂L/∂x = 2 + λ1 = 0
∂L/∂y = 1 + λ1 + 2λ2y = 0
∂L/∂z = λ1 + 2λ2z = 0
∂L/∂λ1 = x + y + z - 1 = 0
∂L/∂λ2 = y^2 + z^2 - 4 = 0
3. Solve the system of equations obtained from step 2 to find the values of x, y, z, λ1, and λ2 at the critical point(s).
4. Substitute the values obtained from step 3 into the objective function (2x + y) to find the minimum value.
Note: The solution using Lagrange multipliers can be computationally intensive, so it's important to check for alternative methods like the one described in the initial explanation.