Lagrange Multipliers

Question

Us‎e the M‎ethod of Lagr‎ange Mul‎tipliers to find the m‎inimum and maxi‎mum val‎ues of the fun‎ction

f(x,y) = 2xy²-2x

wh‎ere the po‎int (x, y) lie‎s on the c‎urve x² + 4y² = 4

Answer 1

Here is a rough sketch. Please check for accuracy and fill in the blanks.

Let's define the Lagrangian as: L(x,y,λ) = f - λ.g = (2xy²-2x) - λ (x² + 4y² - 4) and find its critical points by setting partial derivatives to zero:

1. ∂L/∂x = 2y² - 2 + 2λx = 0
2. ∂L/∂y = 4xy + 8λy = 0
3. ∂L/∂λ = x² + 4y² - 4 = 0

By solving the above equations, we obtain the following critical points: (x=0,y=1), (x=0,y=-1), (x=2,y=0), (x=-2,y=0). Now evaluate the objective f(x,y) = 2xy²-2x at these points to find the min as f(2,0)=-4 and max as f(-2,0)=4.