form-a-Lagrangian-to-maximize-x-2-y-2-subject-to-the-constraint-2x-y-3-

Question Number 95924 by i jagooll last updated on 28/May/20

form a Lagrangian to maximize

x^{2} - y^{2} subject to the

constraint 2 x + y = 3 ?

Commented by john santu last updated on 29/May/20

f (x, y, λ) = x^{2} - y^{2} + λ (2 x + y - 3)

\frac{\partial f}{\partial x} = 2 x + 2 λ = 0 \Rightarrow λ = - x

\frac{\partial f}{\partial y} = - 2 y + λ = 0 \Rightarrow λ = 2 y

\frac{\partial f}{\partial λ} = 2 x + y - 3 = 0 \Rightarrow - 2 λ + \frac{λ}{2} = 3

- 3 λ = 6 \Rightarrow λ = - 2 \Rightarrow {\begin{cases} x = 2 \\ y = - 1 \end{cases}

\max f (2, - 1) = 2^{2} - {(- 1)}^{2} = 3

Answered by mr W last updated on 28/May/20

f (x, y) = x^{2} - y^{2}

F (x, y, λ) = x^{2} - y^{2} + λ (2 x + y - 3)

\frac{\partial F}{\partial x} = 2 x + 2 λ = 0 \Rightarrow x = - λ

\frac{\partial F}{\partial y} = - 2 y + λ = 0 \Rightarrow y = \frac{λ}{2}

\frac{\partial F}{\partial λ} = 2 x + y - 3 = 0 \Rightarrow - 2 λ + \frac{λ}{2} - 3 = 0 \Rightarrow λ = - 2

\Rightarrow x = 2, y = - 1

f_{m a x} = f (2, - 1) = 2^{2} - {(- 1)}^{2} = 3

o r

f (x) = x^{2} - {(3 - 2 x)}^{2}

\frac{d f}{d x} = 2 x - 2 (3 - 2 x) (- 2) = 0

\Rightarrow x = 2

f_{m a x} = f (2) = 2^{2} - {(3 - 4)}^{2} = 3

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