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Question Number 140367 by aliibrahim1 last updated on 06/May/21

	Answered by meetbhavsar25 last updated on 06/May/21

	$A n s w e r i s \frac{9}{25} .$ $x^{2} h a s m i n i m u m v a l u e 0 .$ $y^{2} h a s m i n i m u m v a l u e 0 .$ ${(z - 1)}^{2} h a s m i n i m u m v a l u e 0 . . . b u t i f w e t a k e s o t h e n e q u a t i o n 2 x - 4 y + 5 z = 2 i s n o t s a t i s f y i n g .$ $s o w e h a v e t o t a k e$ $x = 0$ $y = 0$ $z = \frac{2}{5}$ $a n d w e g e t x^{2} + y^{2} + {(z - 1)}^{2} = \frac{9}{25} .$
	Commented by aliibrahim1 last updated on 06/May/21

	$t h x d i r a p p r e c i a t e i t$
	Commented by aliibrahim1 last updated on 06/May/21

	$y o u r a n s w e r i s i n c o r r e c t s i r$
	Commented by meetbhavsar25 last updated on 07/May/21

	$W h a t i s t h e a n s w e r ?$
	Commented by meetbhavsar25 last updated on 07/May/21

	$O k a y . . . . i g o t t h e a n s w e r \frac{1}{5}$
	Commented by aliibrahim1 last updated on 07/May/21

	$y e s c o r r e c t$
	Answered by mr W last updated on 06/May/21

	$z = \frac{2 - 2 x + 4 y}{5}$ $F = x^{2} + y^{2} + {(z - 1)}^{2}$ $= x^{2} + y^{2} + {(\frac{2 - 2 x + 4 y}{5} - 1)}^{2}$ $= \frac{1}{25} (29 x^{2} + 41 y^{2} - 16 x y + 12 x - 24 y + 9)$ $\frac{\partial F}{\partial x} = 0 \Rightarrow 29 x - 8 y = - 6$ $\frac{\partial F}{\partial y} = 0 \Rightarrow - 8 x + 41 y = 12$ $29 \times 41 x - 8 \times 41 y = - 6 \times 41$ $- 8 \times 8 x + 41 \times 8 y = 12 \times 8$ $x = \frac{12 \times 8 - 6 \times 41}{29 \times 41 - 8 \times 8} = - \frac{2}{15}$ $y = \frac{12 \times 29 - 6 \times 8}{41 \times 29 - 8 \times 8} = \frac{4}{15}$ $F_{m i n} = {(- \frac{2}{15})}^{2} + {(\frac{4}{15})}^{2} + {(\frac{- 2 \times \frac{2}{15} - 4 \times \frac{4}{15} + 3}{5})}^{2}$ $= \frac{1}{5}$
	Commented by mr W last updated on 06/May/21

	$a n o t h e r w a y w i t h o u t c a l c u l u s :$ $F = \frac{1}{25} (29 x^{2} + 41 y^{2} - 16 x y + 12 x - 24 y + 9)$ $s a y 29 x^{2} + 41 y^{2} - 16 x y + 12 x - 24 y + 9 = k$ $29 x^{2} + (12 - 16 y) x + 41 y^{2} - 24 y + 9 - k = 0$ $Δ = {(12 - 16 y)}^{2} - 4 \times 29 \times (41 y^{2} - 24 y + 9 - k) ⩾ 0$ ${(6 - 8 y)}^{2} - 29 \times (41 y^{2} - 24 y + 9 - k) ⩾ 0$ $1125 y^{2} - 600 y + 225 - 29 k ⩽ 0$ $Δ = 600^{2} - 4 \times 1125 \times (225 - 29 k) ⩾ 0$ $300^{2} - 1125 \times 225 + 1125 \times 29 k ⩾ 0$ $k ⩾ 5$ $F ⩾ \frac{1}{25} \times 5 = \frac{1}{5}$ $i . e . F_{m i n} = \frac{1}{5}$
	Commented by aliibrahim1 last updated on 07/May/21

	$t h x s i r r e a l l y a p p r e c i a t e i t$
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