Minimize S=x+2y with xy=2 and both x and y > 0
Minimize S=x+2y with xy=2 and both x and y > 0
Put F(x,y) = xy
Then we have:
dS/dx = lambda dF/dx (1)
dS/dy = lambda dF/dy (2)
Eq. (1) gives:
lambda y = 1 (3)
Eq. (2) gives:
lambda x = 2 (4)
If you divide (4) by (3 you get:
x/y = 2 (5)
We also know that
xy=2 (6)
If you multiply (5) by (6) you get x^2 = 4 which yields x = 2 because we knbow that x has to be positive. And this implies that y = 1
To minimize the function S = x + 2y subject to the constraint xy = 2, and with both x and y being greater than 0, we can use the method of Lagrange multipliers.
Step 1: Formulate the Lagrangian function.
The Lagrangian function is given by:
L(x, y, λ) = x + 2y - λ(xy - 2)
Here, λ is the Lagrange multiplier.
Step 2: Find the partial derivatives.
Calculate the partial derivatives of L with respect to x, y, and λ:
∂L/∂x = 1 - λy
∂L/∂y = 2 - λx
∂L/∂λ = xy - 2
Step 3: Set the partial derivatives equal to zero.
Set each partial derivative equal to zero and solve the resulting equations simultaneously:
1 - λy = 0
2 - λx = 0
xy - 2 = 0
Step 4: Solve the equations.
From the equation 1 - λy = 0, solve for λ:
λ = 1/y
Plug this value of λ into the equation 2 - λx = 0:
2 - (1/y)x = 0
2 = (1/y)x
x = 2y
Substitute this value of x into the equation xy - 2 = 0:
(2y)y - 2 = 0
2y^2 - 2 = 0
y^2 - 1 = 0
(y - 1)(y + 1) = 0
y = 1 or y = -1
Since y>0, we reject the solution y = -1.
If y = 1, then x = 2y = 2(1) = 2.
Therefore, the solution that minimizes S is x = 2 and y = 1, with the value of S = x + 2y = 2 + 2(1) = 4.