How-do-I-use-the-Lagrange-multiplier-method-to-find-the-maximum-of-the-function-f-x-y-4-2x-2y-on-the-curve-x-2-y-2-4-

Question Number 137741 by bemath last updated on 06/Apr/21

How do I use the Lagrange multiplier method to find the maximum of the function f(x,y)=4-2x-2y on the curve x^2+y^2=4?

Answered by TheSupreme last updated on 06/Apr/21

y o u c a n f i n d a b s m a x a n d m i n, s e a r c h i n g r e l a t i v e m a x a n d m k n [o f

g (x, y, λ) = 4 - 2 x - 2 y + λ (x^{2} + y^{2} - 4)

s o

g r a d g (x, y, λ) = {\begin{cases} - 2 + 2 λ x = 0 \\ - 2 + 2 λ y = 0 \\ x^{2} + y^{2} - 4 = 0 \end{cases}

{\begin{cases} x = \frac{1}{λ} \\ y = \frac{1}{λ} \\ \frac{2}{λ^{2}} = 4 \end{cases}

(x, y, λ) = (\sqrt{2}, \sqrt{2}, \frac{\sqrt{2}}{2})

(x, y, λ) = (- \sqrt{2}, - \sqrt{2}, - \frac{\sqrt{2}}{2})

Answered by mr W last updated on 06/Apr/21

x^{2} + y^{2} = 2^{2}

\Rightarrow x = 2 \cos θ, y = 2 \sin θ

f (x, y) = 4 - 2 (x + y) = 4 - 4 (\cos θ + \sin θ)

= 4 - 4 \sqrt{2} \sin (θ + \frac{π}{4})

m a x = 4 + 4 \sqrt{2} = 4 (1 + \sqrt{2})

m i n = 4 - 4 \sqrt{2} = 4 (1 - \sqrt{2})

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