Math Question and Answers


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 44384 by Joel578 last updated on 28/Sep/18

$Let a and b are real numbers such that$ $a > b > 0$ $Find the minimum value of$ $\sqrt{2} a^{3} + \frac{3}{a b - b^{2}}$
Commented bybehi83417@gmail.com last updated on 28/Sep/18

$f (a, b) = \sqrt{2} a^{3} + \frac{3}{a b - b^{2}}$ $\frac{\partial f}{\partial a} = 3 \sqrt{2} a^{2} - \frac{3 b}{{(a b - b^{2})}^{2}} = 0 (i)$ $\frac{\partial f}{\partial b} = - \frac{3 (a - 2 b)}{{(a b - b^{2})}^{2}} = 0 (i i)$ $(i i) \Rightarrow a - 2 b = 0 \Rightarrow a = 2 b$ $(i) \Rightarrow 12 \sqrt{2} b^{2} - \frac{3 b}{{(2 b^{2} - b^{2})}^{2}} = 0 \Rightarrow$ $12 \sqrt{2} b^{2} - \frac{3}{b^{3}} = 0 \Rightarrow b^{5} = \frac{1}{4 \sqrt{2}} = \frac{1}{2^{\frac{5}{2}}} = \frac{1}{{(\sqrt{2})}^{5}}$ $\Rightarrow b = \frac{\sqrt{2}}{2}, a = \sqrt{2}$ $\Rightarrow f_{m i n} = \sqrt{2} {(\sqrt{2})}^{3} + \frac{3}{\sqrt{2} . \frac{\sqrt{2}}{2} - \frac{1}{2}} =$ $= 4 + \frac{3}{1 - \frac{1}{2}} = 10 .$
Commented byajfour last updated on 28/Sep/18

$V e r y g r a t e f u l S i r!$
Commented byrahul 19 last updated on 28/Sep/18

$T h a n k s t o {M r w}_{3} s i r f o r s h a r i n g t h i s$ $m e t h o d w i t h u s!!$
Commented byJoel578 last updated on 28/Sep/18

$Thank you very much$
Commented bytanmay.chaudhury50@gmail.com last updated on 28/Sep/18

Commented bytanmay.chaudhury50@gmail.com last updated on 28/Sep/18

	Answered by ajfour last updated on 28/Sep/18

	$l e t x = \sqrt{2} a^{3} + \frac{3}{b (a - b)}$ $\Rightarrow b (a - b) = \frac{3}{(x - \sqrt{2} a^{3})}$ $i f x i s m i n i m u m, b (a - b) i s m a x i m u m$ $\Rightarrow b = \frac{a}{2}$ $\Rightarrow x - \sqrt{2} a^{3} = \frac{12}{a^{2}}$ $x = \sqrt{2} a^{3} + \frac{12}{a^{2}}$ $w h i c h h a s a m i n i m u m v a l u e$ $w h e n 3 \sqrt{2} a^{2} = \frac{24}{a^{3}}$ $\Rightarrow a^{5} = {(\sqrt{2})}^{5} \Rightarrow a = \sqrt{2}$ $\Rightarrow x_{m i n} = 4 + \frac{12}{2} = 10 .$
	Commented byJoel578 last updated on 28/Sep/18

	$Thank you very much$
	Answered by tanmay.chaudhury50@gmail.com last updated on 28/Sep/18
	$f(a,b)=(√2) a^3 +3(ab−b^2 )^(−1) (∂f/∂a)=3(√2) a^2 −3(ab−b^2 )^(−2) (b) (∂f/∂a)=0=3(√2) a^2 −((3b)/((ab−b^2 )^2 )) we get a=(√2) (∂^2 f/∂a^2 )=6(√2) a+6(ab−b^2 )^(−3) (b)^2 (∂f/∂b)=0−3(ab−b^2 )^(−2) (a−2b) (∂f/∂b)=0=((−3)/((ab−b^2 )^2 ))(a−2b) we get b=(1/((√2) )) a=2b 3(√2)a^2 −((3b)/((ab−b^2 )^2 ))=0 12(√2) b^2 −((3b)/b^4 )=0 12(√2) b^6 −3b=0 3b(4(√2) b^5 −1)=0 b=((1/(4(√2))))^(1/5) ={((1/(√2)) )^5 }^(1/5) =(1/(√2)) a=2×(1/(√2))=(√2) (∂^2 b/∂b^2 )=6(ab−b^2 )^(−3) (a−2b)^2 −3(ab−b^2 )(0−2) now putting the value in (∂^2 f/∂a^2 ) and (∂^2 f/∂b^2 ) (∂^2 f/∂a^2 )=6(√2) a+((6b^2 )/((ab−b^2 )^3 )) =6(√2) ×(√2) +((6×(1/2))/((1−(1/2))^3 ))=12+(3/(1/8))=36 (∂^2 f/∂b^2 )=6(1−(1/2))^(−3) ((√2) −(2/((√2) )))^2 +6(1−(1/2))=3 (∂/∂a)((∂f/∂b))=(∂/∂a){((6b−3a)/((ab−b^2 )^2 ))} =(((ab−b^2 )^2 (−3)−(6b−3a)×2(ab−b^2 )(b))/((ab−b^2 )^4 )) =(((1−(1/2))^2 (−3)−((6/(√2))−3(√2) )×(1−(1/2))((1/2)))/((1−(1/2))^4 )) =(((−3)/4)/(1/(16)))=−12 (∂^2 f/∂a^2 )×(∂^2 f/∂b^2 )−{(∂/∂a)((∂f/∂b))}^2 =36×3−(−12)^2 =108−144=−36<0 saddle point...$
	$f (a, b) = \sqrt{2} a^{3} + 3 {(a b - b^{2})}^{- 1}$ $\frac{\partial f}{\partial a} = 3 \sqrt{2} a^{2} - 3 {(a b - b^{2})}^{- 2} (b)$ $\frac{\partial f}{\partial a} = 0 = 3 \sqrt{2} a^{2} - \frac{3 b}{{(a b - b^{2})}^{2}} w e g e t a = \sqrt{2}$ $\frac{\partial^{2} f}{\partial a^{2}} = 6 \sqrt{2} a + 6 {(a b - b^{2})}^{- 3} {(b)}^{2}$ $\frac{\partial f}{\partial b} = 0 - 3 {(a b - b^{2})}^{- 2} (a - 2 b)$ $\frac{\partial f}{\partial b} = 0 = \frac{- 3}{{(a b - b^{2})}^{2}} (a - 2 b) w e g e t b = \frac{1}{\sqrt{2}}$ $a = 2 b$ $3 \sqrt{2} a^{2} - \frac{3 b}{{(a b - b^{2})}^{2}} = 0$ $12 \sqrt{2} b^{2} - \frac{3 b}{b^{4}} = 0$ $12 \sqrt{2} b^{6} - 3 b = 0$ $3 b (4 \sqrt{2} b^{5} - 1) = 0$ $b = {(\frac{1}{4 \sqrt{2}})}^{\frac{1}{5}} = {{(\frac{1}{\sqrt{2}})}^{5}}^{\frac{1}{5}} = \frac{1}{\sqrt{2}}$ $a = 2 \times \frac{1}{\sqrt{2}} = \sqrt{2}$ $\frac{\partial^{2} b}{\partial b^{2}} = 6 {(a b - b^{2})}^{- 3} {(a - 2 b)}^{2} - 3 (a b - b^{2}) (0 - 2)$ $n o w p u t t i n g t h e v a l u e i n \frac{\partial^{2} f}{\partial a^{2}} a n d \frac{\partial^{2} f}{\partial b^{2}}$ $\frac{\partial^{2} f}{\partial a^{2}} = 6 \sqrt{2} a + \frac{6 b^{2}}{{(a b - b^{2})}^{3}}$ $= 6 \sqrt{2} \times \sqrt{2} + \frac{6 \times \frac{1}{2}}{{(1 - \frac{1}{2})}^{3}} = 12 + \frac{3}{\frac{1}{8}} = 36$ $\frac{\partial^{2} f}{\partial b^{2}} = 6 {(1 - \frac{1}{2})}^{- 3} {(\sqrt{2} - \frac{2}{\sqrt{2}})}^{2} + 6 (1 - \frac{1}{2}) = 3$ $\frac{\partial}{\partial a} (\frac{\partial f}{\partial b}) = \frac{\partial}{\partial a} {\frac{6 b - 3 a}{{(a b - b^{2})}^{2}}}$ $= \frac{{(a b - b^{2})}^{2} (- 3) - (6 b - 3 a) \times 2 (a b - b^{2}) (b)}{{(a b - b^{2})}^{4}}$ $= \frac{{(1 - \frac{1}{2})}^{2} (- 3) - (\frac{6}{\sqrt{2}} - 3 \sqrt{2}) \times (1 - \frac{1}{2}) (\frac{1}{2})}{{(1 - \frac{1}{2})}^{4}}$ $= \frac{\frac{- 3}{4}}{\frac{1}{16}} = - 12$ $\frac{\partial^{2} f}{\partial a^{2}} \times \frac{\partial^{2} f}{\partial b^{2}} - {\frac{\partial}{\partial a} (\frac{\partial f}{\partial b})}^{2}$ $= 36 \times 3 - {(- 12)}^{2}$ $= 108 - 144 = - 36 < 0 s a d d l e p o i n t . . .$
Terms of Service
Privacy Policy
Contact: info@tinkutara.com

Question and Answers Forum