Let x,y∈R such that 2x^2 +3y^2 =5 Find the minimum and maximum of expression: P=x^3 −y^3 +x−2y


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Question Number 173707 by dragan91 last updated on 16/Jul/22

$Let x, y \in R such that {2 x}^{2} + {3 y}^{2} = 5$ $Find the minimum and$ $maximum of expression :$ $P = x^{3} - y^{3} + x - 2 y$
	Answered by a.lgnaoui last updated on 17/Jul/22

	$\frac{d (P)}{d x} + \frac{d (P)}{d y} = 3 x^{2} + 1 - 3 y^{2} - 2$ $= 3 x^{2} - 3 y^{2} - 1 3 y^{2} = 5 - 2 x^{2}$ $3 x^{2} - 1 - (5 - 2 x^{2}) = 5 x^{2} - 6$ $t h e m a x i m u m a n d m i n i m u m o f P v e r i f y P^{'} (x) = 0$ $5 x^{2} - 6 = 0 x = \pm \sqrt{\frac{6}{5}}$ $x = \pm \sqrt{\frac{6}{5}} y = \sqrt{(5 - 2 \frac{6}{5}) / 3} = \sqrt{\frac{13}{15}}$ $m = (- \sqrt{\frac{6}{5},} \sqrt{\frac{13}{15}}) m a x = (+ \sqrt{\frac{6}{5}}, \sqrt{\frac{13}{15}})$
	Commented by mr W last updated on 17/Jul/22

	$i t h i n k y o u r m e t h o d i s w r o n g .$ $f o r c o n d i t i o n a l e x t r e m u m s i t {d o e s n}^{'} t$ $h o l d t h a t \frac{\partial P}{\partial x} = 0 \land \frac{\partial P}{\partial y} = 0, o r \frac{\partial P}{\partial x} + \frac{\partial P}{\partial y} = 0$ $a s y o u s a i d .$
	Answered by mr W last updated on 17/Jul/22

	$G = x^{3} - y^{3} + x - 2 y + λ (2 x^{2} + 3 y^{2} - 5)$ $\frac{\partial G}{\partial x} = 3 x^{2} + 1 + 4 λ x = 0 \Rightarrow x = \frac{- 2 λ \pm \sqrt{4 λ^{2} - 3}}{3}$ $\frac{\partial G}{\partial y} = - 3 y^{2} - 2 + 6 λ y = 0 \Rightarrow y = \frac{3 λ \pm \sqrt{9 λ^{2} - 6}}{3}$ $\frac{\partial G}{\partial λ} = 2 x^{2} + 3 y^{2} - 5 = 0$ $\Rightarrow 2 {[\frac{- 2 λ \pm \sqrt{4 λ^{2} - 3}}{3}]}^{2} + 3 {[\frac{3 λ \pm \sqrt{9 λ^{2} - 6}}{3}]}^{2} - 5 = 0$ ${t h a t}^{'} s 4 e q u a t i o n s w h i c h a l l {c a n}^{'} t$ $b e s o l v e d e x a c t l y .$ $w e g e t a p p r o x i m a t e l y$ $λ = - 1.3191, P = 5.8033 = m a x$ $λ = 0.8872, P = - 4.9111$ $λ = - 0.8872, P = 4.9111$ $λ = 1.3191, P = - 5.8033 = m i n$
	Commented by mr W last updated on 17/Jul/22

	Commented by Tawa11 last updated on 17/Jul/22

	$Great sir$
	Commented by a.lgnaoui last updated on 17/Jul/22

	$t h a n k y o u p r o f e s s o r$
	Answered by MJS_new last updated on 17/Jul/22

	$2 x^{2} + 3 y^{2} = 5 \Rightarrow y = σ \frac{\sqrt{3 (5 - 2 x^{2})}}{3} \land σ = \pm 1$ $P (x) = x (x^{2} + 1) + σ \frac{(2 x^{2} - 11) \sqrt{3 (5 - 2 x^{2})}}{9}$ $P^{'} (x) = 0$ $3 x^{2} + 1 - σ \frac{2 \sqrt{3} x (2 x^{2} - 7)}{3 \sqrt{5 - 2 x^{2}}} = 0$ $\Rightarrow$ $x^{6} - \frac{211}{70} x^{4} + \frac{8}{5} x^{2} - \frac{3}{14} = 0$ $t = x^{2}$ $t^{3} - \frac{211}{70} t^{2} + \frac{8}{5} t - \frac{3}{14} = 0$ $good luck! t = \frac{3}{14} is a solution$ $t = \frac{3}{14} \lor t = \frac{7}{5} \pm \frac{2 \sqrt{6}}{5}$ $now {it}^{'} s easy to get$ $m i n (P (x)) = - \frac{(9 + 44 \sqrt{6}) \sqrt{5}}{45} at x = - \frac{(1 + \sqrt{6}) \sqrt{5}}{5} \land y = \frac{(3 - 2 \sqrt{6}) \sqrt{5}}{15}$ $m a x (P (x)) = \frac{(9 + 44 \sqrt{6}) \sqrt{5}}{45} at x = \frac{(1 + \sqrt{6}) \sqrt{5}}{5} \land y = - \frac{(3 - 2 \sqrt{6}) \sqrt{5}}{15}$ $[approximated values \approx \pm 5.80272 at x \approx \pm 1.54266 \land y \approx \pm . 283083]$
	Commented by Tawa11 last updated on 17/Jul/22

	$Great sir$
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