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Question Number 44230 by behi83417@gmail.com last updated on 24/Sep/18

	Answered by MrW3 last updated on 25/Sep/18

	$(1)$ $f (x, y) = x^{2} + x y + y^{2} - a x - b y$ $\frac{\partial f}{\partial x} = 2 x + y - a = 0 . . . (i)$ $\frac{\partial f}{\partial y} = x + 2 y - b = 0 . . . (i i)$ $3 x - 2 a + b = 0$ $x = \frac{2 a - b}{3}$ $3 y - 2 b + a = 0$ $y = \frac{2 b - a}{3}$ $f_{m i n} = {(\frac{2 a - b}{3})}^{2} + (\frac{2 a - b}{3}) (\frac{2 b - a}{3}) + {(\frac{2 b - a}{3})}^{2} - a (\frac{2 a - b}{3}) - b (\frac{2 b - a}{3})$ $= \frac{1}{9} [4 a^{2} - 4 a b + b^{2} + 4 a b - 2 b^{2} - 2 a^{2} + a b + 4 b^{2} - 4 a b + a^{2} - 6 a^{2} + 3 a b - 6 b^{2} + 3 a b]$ $= \frac{1}{3} (a b - a^{2} - b^{2})$ $(2)$ $g (x, y) = \frac{{(a x + b y + c)}^{2}}{x^{2} + y^{2} + 1}$ $\frac{\partial g}{\partial x} = - \frac{2 x {(a x + b y + c)}^{2}}{{(x^{2} + y^{2} + 1)}^{2}} + \frac{2 a (a x + b y + c)}{x^{2} + y^{2} + 1} = 0$ $x (a x + b y + c) = a (x^{2} + y^{2} + 1) . . . (i)$ $\frac{\partial g}{\partial y} = - \frac{2 y {(a x + b y + c)}^{2}}{{(x^{2} + y^{2} + 1)}^{2}} + \frac{2 b (a x + b y + c)}{x^{2} + y^{2} + 1} = 0$ $y (a x + b y + c) = b (x^{2} + y^{2} + 1) . . . (i i)$ $\frac{x}{a} = \frac{y}{b}$ $x (a x + \frac{b^{2} x}{a} + c) = a (x^{2} + \frac{b^{2} x^{2}}{a^{2}} + 1)$ $a x (a^{2} x + b^{2} x + a c) = a (a^{2} x^{2} + b^{2} x^{2} + a^{2})$ $\Rightarrow x = \frac{a}{c}$ $\Rightarrow y = \frac{b}{c}$ $\Rightarrow g_{m a x} = \frac{{(\frac{a^{2}}{c} + \frac{b^{2}}{c} + c)}^{2}}{\frac{a^{2}}{c^{2}} + \frac{b^{2}}{c^{2}} + 1} = a^{2} + b^{2} + c^{2}$
	Commented by behi83417@gmail.com last updated on 25/Sep/18

	$t h a n k y o u d e a r m a s t e r .$ $n i c e w o r k d o n e b y y o u . p e r f e c t!$
	Commented by tanmay.chaudhury50@gmail.com last updated on 25/Sep/18

	$b a h . . . d a r u n . . . e x c e l l e n t . . .$
	Commented by MrW3 last updated on 25/Sep/18

	$t h a n k y o u b o t h s i r s!$
	Commented by rahul 19 last updated on 26/Sep/18

	$S i r, h o w t o c h e c k w h e t h e r i t w i l l b e$ $g_{m a x} / g_{m i n} a n d f_{m i n} / f_{m a x} ?$
	Commented by MrW3 last updated on 26/Sep/18

	$w e c h e c k i f \frac{\partial^{2} f}{\partial x^{2}} a n d \frac{\partial^{2} f}{\partial y^{2}} a r e + v e o r - v e .$
	Commented by rahul 19 last updated on 26/Sep/18

	$S u p p o s e t h e r e i s a c a s e w h e r e$ $\frac{\partial^{2} f}{\partial x^{2}} = + v e a n d \frac{\partial^{2} f}{\partial y^{2}} = - v e .$ $T h e n W h a t w i l l i s a y ?$
	Commented by MrW3 last updated on 26/Sep/18

	$m a x . i n y - d i r e c t i o n b u t$ $m i n . i n x - d i r e c t i o n$ $\Rightarrow n e i t h e r a m a x . n o r a m i n . f o r$ $t h e f u n c t i o n$
	Commented by rahul 19 last updated on 26/Sep/18

	$O k s i r, t h a n k s!$
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