Answer:
- The values of x and y that minimize the function, subject to the given constraint are 6 and 8 respectively.
- The minimum value of the function = -44
Step-by-step explanation:
The Lagrange multiploer method finds the optimum value of a multivariable function subjected to a given constraint
It replaces the function with a Lagrange equivalent which is
L(x, y) = F(x, y) - λ C(x, y)
where λ Is the lagrange multiplier which can be a function of x and y
F(x, y) = x² - 10x + y² - 14y + 28
C(x, y) = x + y - 14
L(x, y) = x² - 10x + y² - 14y + 28 - λ (x + y - 14)
We now take the partial derivatives of the Lagrange function with respect to x, y and λ respecrively. Then solving to obtain values of x, y and λ that correspond to the minimum of the function. Since the first partial derivatives are all equal to 0 at minimum point.
(∂L/∂x) = 2x - 10 - λ = 0 (eqn 1)
(∂L/∂y) = 2y - 14 - λ = 0 (eqn 2)
(∂L/∂λ) = x + y - 14 = 0 (eqn 3)
Equating eqn 1 and 2
2x - 10 - λ = 2y - 14 - λ
2x - 10 = 2y - 14
2y = 2x - 10 + 14
2y = 2x + 4
y = x + 2 (eqn *)
Substitute eqn ^ into eqn 3
x + y - 14 = 0
x + x + 2 - 14 = 0
2x - 12 = 0
2x = 12
x = 6
y = x + 2 = 6 + 2 = 8
2x - 10 - λ = 0
12 - 10 - λ = 0
λ = 2
The values of x and y that minimize the function are 6 and 8 respectively.
F(x, y) = x² - 10x + y² - 14y + 28
At minimum point, x = 6, y = 8
F(x, y) = 6² - 10(6) + 8² - 14(8) + 28 = 36 - 60 + 64 - 112 + 28 = -44
Hope this Helps!!!
