if-x-2-y-2-50-find-the-minimum-and-maximum-of-x-y-2-8-x-y-20-

Question Number 79190 by mr W last updated on 23/Jan/20

i f x^{2} + y^{2} = 50,

f i n d t h e m i n i m u m a n d m a x i m u m o f

{(x + y)}^{2} - 8 (x + y) + 20

Commented by jagoll last updated on 23/Jan/20

\Rightarrow x^{2} + y^{2} + 2 xy - 8 x - 8 y + 20 =

2 xy - 8 x - 8 y + 70

let f (x, y, λ) = 2 xy - 8 x - 8 y + 70 + λ (x^{2} + y^{2} - 50)

Commented by jagoll last updated on 23/Jan/20

we use differential

Commented by jagoll last updated on 23/Jan/20

right sir ?

Commented by jagoll last updated on 23/Jan/20

\frac{df}{dx} = 2 y - 8 + λ (2 x) = 0

\frac{df}{dy} = 2 x - 8 + λ (2 y) = 0

\frac{df}{d λ} = x^{2} + y^{2} - 50 = 0

Commented by mind is power last updated on 23/Jan/20

n i c e B u t y o u r a p p r o c h {d i d n}^{'} t u s e t h e c o n t r a i n t x^{2} + y^{2} = 50

t r y t o p u t g (x, y, γ) = 2 x y - 8 x - 8 y + 70 + γ (x^{2} + y^{2} - 50)

Commented by jagoll last updated on 23/Jan/20

oo yes . it Langrange method

Commented by jagoll last updated on 23/Jan/20

thank you sir

Commented by jagoll last updated on 23/Jan/20

that right sir ?

Commented by jagoll last updated on 23/Jan/20

\frac{2 λ x}{2 λ y} = \frac{8 - 2 y}{8 - 2 x} \Rightarrow \frac{x}{y} = \frac{4 - y}{4 - x}

4 x - x^{2} = 4 y - y^{2}

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

x - 2 = \pm (y - 2) \Rightarrow x = 2 \pm (y - 2)

x_{1} = y \lor x_{2} = 4 - y

Commented by jagoll last updated on 23/Jan/20

case (1) x^{2} + y^{2} = 50 \Rightarrow x = \pm 5 = y

f = {(10)}^{2} - 8 (10) + 20 = 40

Commented by jagoll last updated on 23/Jan/20

case (2) {(4 - y)}^{2} + y^{2} - 50 = 0

{2 y}^{2} - 8 y - 34 = 0

y^{2} - 4 y - 17 = 0

{(y - 2)}^{2} - 21 = 0 \Rightarrow y = 2 + \sqrt{21}

Commented by jagoll last updated on 23/Jan/20

mr W what the answer ?

Commented by mind is power last updated on 23/Jan/20

N i c e S i r

Answered by mind is power last updated on 23/Jan/20

x = \sqrt{50} c o s (θ)

y = \sqrt{50} s i n (θ)

x + y = \sqrt{50} . (c o s (θ) + s i n (θ))

= 10 s i n (θ + \frac{π}{4}) = 10 s i n (φ), φ \in [0, 2 π [

x + y \in [- 10, 10]

{(x + y)}^{2} - 8 (x + y) + 20 = f (x + y)

f (t) = t^{2} - 8 t + 20

t \in [- 10, 10]

f^{'} (t) = 2 t - 8, m i n f = f (4) = 16 - 32 + 20 = 4

x + y = 4 = 10 s i n (φ), φ = \sin^{- 1} (\frac{2}{5}), θ = \sin^{- 1} (\frac{2}{5}) - \frac{π}{4}

x = \sqrt{50} c o s (θ), y = \sqrt{50} s i n (θ)

m o x f (- 10) = 100 + 80 + 20 = 200

- 10 = 10 s i n (φ) \Rightarrow φ = - \frac{π}{2} \Rightarrow θ = - \frac{3 π}{4}

x = \sqrt{50} . - \frac{\sqrt{2}}{2} = - 5, y = - 5

Answered by mr W last updated on 23/Jan/20

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

m a x i m u m o f f = {(x + y)}^{2} - 8 (x + y) + 20

u n d e r t h e c o n d i t i o n t h a t x, y s a t i s f y

x^{2} + y^{2} = 50, i . e . (x, y) i s o n t h e c i r c l e .

l e t x = 5 \sqrt{2} \cos θ, y = 5 \sqrt{2} \sin θ

f = {(x + y)}^{2} - 8 (x + y) + 20

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

= {(5 \sqrt{2} \cos θ + 5 \sqrt{2} \sin θ - 4)}^{2} + 4

= {[10 (\frac{1}{\sqrt{2}} \cos θ + \frac{1}{\sqrt{2}} \sin θ) - 4]}^{2} + 4

= {[10 \cos (θ - \frac{π}{4}) - 4]}^{2} + 4 ⩾ 4

f_{m i n} i s w h e n \cos (θ - \frac{π}{4}) = \frac{4}{10}, i . e . θ = \frac{π}{4} + \cos^{- 1} \frac{2}{5},

f_{m i n} = {(4 - 4)}^{2} + 4 = 4

f_{m a x} i s w h e n \cos (θ - \frac{π}{4}) = - 1, i . e . θ = \frac{5 π}{4},

f_{m a x} = {(- 10 - 4)}^{2} + 4 = 200

Answered by key of knowledge last updated on 23/Jan/20

u = {(x + y)}^{2} - 8 (x + y) + 20 = {((x + y) - 4)}^{2} + 4

if ((x + y) - 4) = 0 \Rightarrow \min (u) = 4

\min (x + y) = - 10 & \max (x + y) = 10

ax ((x + y) - 4) = (- 14) \Rightarrow \max (u) = 200

Answered by john santu last updated on 24/Jan/20

b e c a u s e x a n d y m e e t t h e

c o n s t r a i n t s o f c i r c l e x^{2} + y^{2} = 50

, l e t x + y = k i s t a n g e n t l i n e c i r c l e,

t h e n d i s t a n c e o f t h e c e n t e r

o f t h e c i r c l e t o t a n g e n t i s t h e

r a d i u s \Rightarrow r =∣ \frac{k}{\sqrt{2}} ∣, ∣ k ∣= 10 \Rightarrow k = \pm 10

f o r x + y = - 10 \Rightarrow f = {(- 10)}^{2} - 8 (- 10) + 20 = 200

∴ f_{m a x} = 200

c o n s i d e r x + y = t

f (t) = t^{2} - 8 t + 20, a q u a d r a t i c

f u n c t i o n i n t . t h e m i n i m u m

v a l u e w h e n t = \frac{- b}{2 a} = \frac{- (- 8)}{2.1} = 4

t h e f_{m i n} = 4^{2} - 8 \times 4 + 20 = 4

Commented by mr W last updated on 23/Jan/20

f_{m i n} \neq 40 b u t = 4 .

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