find-maximum-value-2x-2-y-2-with-constraint-x-2-y-2-4x-2y-1-0-

Question Number 87989 by john santu last updated on 07/Apr/20

find maximum value

{2 x}^{2} + y^{2} with constraint

x^{2} + y^{2} - 4 x + 2 y + 1 = 0

Answered by mr W last updated on 07/Apr/20

x^{2} + y^{2} - 4 x + 2 y + 1 = 0

\Rightarrow {(x - 2)}^{2} + {(y + 1)}^{2} = 4

l e t x = 2 + 2 \cos θ

l e t y = - 1 + 2 \sin θ

k = 2 x^{2} + y^{2} = 8 {(1 + \cos θ)}^{2} + {(- 1 + 2 \sin θ)}^{2}

\frac{d k}{d θ} = 16 (1 + \cos θ) (- \sin θ) + 4 (- 1 + 2 \sin θ) \cos θ = 0

4 \sin θ + 2 \cos θ \sin θ + \cos θ = 0

\Rightarrow \frac{4}{\cos θ} + 2 + \frac{1}{\sin θ} = 0

\Rightarrow θ \approx - 0.1659, 2.7118

k_{m i n} \approx 0.094 w i t h θ = 2.7118

k_{m a x} \approx 33.33 w i t h θ = - 0.1659

Commented by mr W last updated on 07/Apr/20

t h e q u e s t i o n i s t o f i n d t h e s m a l l e s t

a n d t h e l a r g e s t e l l i p s e w h i c h t o u c h e s

a g i v e n c i r c l e . “ {e x a c t}^{″} s o l u t i o n m a y b e

n o t p o s s i b l e .

Answered by john santu last updated on 08/Apr/20

Commented by mr W last updated on 08/Apr/20

w h e n i u s e t = \tan \frac{θ}{2}, i g e t a l s o a

q u a d r a t i c e q u a t i o n a b o u t t :

t^{4} - 4 t^{3} - 12 t + 1 = 0 .

Commented by john santu last updated on 08/Apr/20

y e s . w e c a n n o t g e t t h e e x a c t v a l u e