x,y,z ∈ R { ((xy + yz + zx = 3)),((x + y + z = 5)) :} → max(z) = ?


Question and Answers Forum

All Questions Topic List
Algebra Questions
Previous in All Question Next in All Question
Previous in Algebra Next in Algebra
Question Number 201613 by hardmath last updated on 09/Dec/23

$x, y, z \in R$ ${\begin{cases} xy + yz + zx = 3 \\ x + y + z = 5 \end{cases} \to \max (z) = ?$
	Answered by aleks041103 last updated on 09/Dec/23

	$x + y + z = 5 \Rightarrow z = 5 - x - y$ $\Rightarrow x y + (x + y) (5 - (x + y)) = 3$ $x y + 5 x + 5 y - x^{2} - 2 x y - y^{2} = 3$ $\Rightarrow x^{2} + x y + y^{2} - 5 x - 5 y + 3 = 0 (1)$ $m a x z \Leftrightarrow d z = 0 = - d x - d y \Rightarrow d x = - d y$ $d (1)$ $2 x d x + y d x + x d y + 2 y d y - 5 d x - 5 d y = 0$ $(2 x + y - 5) d x + (x + 2 y - 5) d y = 0$ $d x = - d y$ $\Rightarrow (2 x + y - 5) - (x + 2 y - 5) = x - y = 0$ $\Rightarrow x = y$ $(1) \Rightarrow x^{2} + x^{2} + x^{2} - 5 x - 5 x + 3 = 0$ $3 x^{2} - 10 x + 3 = 0$ $x_{1, 2} = \frac{10 \pm \sqrt{100 - 4.3.3}}{6} = \frac{5 \pm 4}{3} = \frac{1}{3}; 3$ $\Rightarrow z_{1} = 5 - 2 x_{1} = \frac{13}{3}$ $z_{2} = 5 - 2 x_{2} = - 1$ $\Rightarrow m a x (z) = \frac{13}{3}$
	Commented by aleks041103 last updated on 09/Dec/23

	$t h e g e o m e t r i c i d e a i s :$ $(1) \to e l l i p s e$ $O n a n o t h e r n o t e :$ $z = 5 - (x + y)$ $x + y = (\begin{matrix} x \\ y \end{matrix}) . (\begin{matrix} 1 \\ 1 \end{matrix}) = \sqrt{2} \vec{n} . \vec{r}$ $\Rightarrow x + y = \sqrt{2} \times (d i s t a n c e f r o m t h e l i n e x + y = 0)$ $\Rightarrow z = 5 - (x + y) i s m a x (m i n)$ $w h e n t h e d i s t a n c e o f (x, y) f r o m t h e l i n e$ $x + y = 0 i s m i n (m a x) .$ $\Rightarrow w e w a n t t o f i n d t h e p o i n t s$ $a t w h i c h t h e t a n g e n t s t o t h e e l l i p s e$ $a r e p a r a l l e l t o t h e e l l i p s e .$
	Commented by aleks041103 last updated on 09/Dec/23

	Commented by hardmath last updated on 10/Dec/23

	$perfect dear professor thank you so much$
	Answered by mr W last updated on 10/Dec/23

	$s a y t = x + y$ $z = 5 - t$ $t o f i n d m a x (z), w e o n l y n e e d t o f i n d$ $m i n (t) .$ $x y + (x + y) z = 3$ $x y + (x + y) (5 - (x + y)) = 3$ ${(x + y)}^{2} - 5 (x + y) + 3 = x y ⩽ {(\frac{x + y}{2})}^{2} = \frac{{(x + y)}^{2}}{4}$ $3 {(x + y)}^{2} - 20 (x + y) + 12 ⩽ 0$ $o r 3 t^{2} - 20 t + 12 ⩽ 0$ $(3 t - 2) (t - 6) ⩽ 0$ $\Rightarrow \frac{2}{3} ⩽ t ⩽ 6$ $i . e . m i n (t) = \frac{2}{3}, m a x (t) = 6$ $\Rightarrow m a x (z) = 5 - \frac{2}{3} = \frac{13}{3}$ $\Rightarrow m i n (z) = 5 - 6 = - 1$
	Commented by mr W last updated on 10/Dec/23

	$A . M . ⩾ G . M .$ $\frac{a + b}{2} ⩾ \sqrt{a b} \Rightarrow a b ⩽ {(\frac{a + b}{2})}^{2}$
	Commented by hardmath last updated on 10/Dec/23

	$perfect dear professor thank you so much$
Terms of Service
Privacy Policy
Contact: info@tinkutara.com

Question and Answers Forum