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...

Explain Bot Explain Bot answered
8 months ago

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, 位1, 位2) as follows:
L(x, y, z, 位1, 位2) = 2x + y + 位1(x + y + z - 1) + 位2(y^2 + z^2 - 4)

2. Take partial derivatives of L with respect to x, y, z, 位1, and 位2 and set them equal to zero to find the critical points:

鈭侺/鈭倄 = 2 + 位1 = 0
鈭侺/鈭倅 = 1 + 位1 + 2位2y = 0
鈭侺/鈭倆 = 位1 + 2位2z = 0
鈭侺/鈭偽1 = x + y + z - 1 = 0
鈭侺/鈭偽2 = y^2 + z^2 - 4 = 0

3. Solve the system of equations obtained from step 2 to find the values of x, y, z, 位1, and 位2 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.

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