a+2b+3c=12 2ab+3ac+6bc=48 a+b+c=...


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Question Number 38391 by gunawan last updated on 25/Jun/18

$a + 2 b + 3 c = 12$ $2 a b + 3 a c + 6 b c = 48$ $a + b + c = . . .$
	Answered by MrW3 last updated on 26/Jun/18
	$let x=a y=2b z=3c (i) ⇒x+y+z=12⇒z=12−x−y (ii) ⇒xy+yz+zx=48⇒xy+12(x+y)−(x+y)^2 =48 F(x,y)=xy+12(x+y)−(x+y)^2 (∂F/∂x)=y+12−2(x+y)=0⇒2x+y=12 (∂F/∂y)=x+12−2(x+y)=0⇒x+2y=12 { ((2x+y=12)),((x+2y=12)) :}⇒x=y=4⇒F(4,4)=48 ⇒maximum from F(x,y) is at x=y=4 which is 48, that is to say { ((x+y+z=12)),((xy+yz+zx=48)) :} has one and only one solution: x=y=z=4 ⇒a=4, b=(4/2)=2, c=(4/3) ⇒a+b+c=4+2+(4/3)=((22)/3)$
	$l e t$ $x = a$ $y = 2 b$ $z = 3 c$ $(i) \Rightarrow x + y + z = 12 \Rightarrow z = 12 - x - y$ $(i i) \Rightarrow x y + y z + z x = 48 \Rightarrow x y + 12 (x + y) - {(x + y)}^{2} = 48$ $F (x, y) = x y + 12 (x + y) - {(x + y)}^{2}$ $\frac{\partial F}{\partial x} = y + 12 - 2 (x + y) = 0 \Rightarrow 2 x + y = 12$ $\frac{\partial F}{\partial y} = x + 12 - 2 (x + y) = 0 \Rightarrow x + 2 y = 12$ ${\begin{cases} 2 x + y = 12 \\ x + 2 y = 12 \end{cases} \Rightarrow x = y = 4 \Rightarrow F (4, 4) = 48$ $\Rightarrow m a x i m u m f r o m F (x, y) i s a t x = y = 4$ $w h i c h i s 48, t h a t i s t o s a y$ ${\begin{cases} x + y + z = 12 \\ x y + y z + z x = 48 \end{cases}$ $h a s o n e a n d o n l y o n e s o l u t i o n :$ $x = y = z = 4$ $\Rightarrow a = 4, b = \frac{4}{2} = 2, c = \frac{4}{3}$ $\Rightarrow a + b + c = 4 + 2 + \frac{4}{3} = \frac{22}{3}$
	Commented by tanmay.chaudhury50@gmail.com last updated on 25/Jun/18

	$e x c e l l e n t . . .$
	Commented by gunawan last updated on 26/Jun/18

	$wow nice Sir$
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