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Question Number 169117 by infinityaction last updated on 24/Apr/22
Answered by greougoury555 last updated on 24/Apr/22
f(x,y,λ)= (x^3 +1)(y^3 +1)+λ(x+y−1) (∂f/∂x) = 3x^2 (y^3 +1)+λ=0 (∂f/∂y) = 3y^2 (x^3 +1)+λ=0 (∂f/∂λ) = x+y=1 ⇒3x^2 (y^3 +1)=3y^2 (x^3 +1) ⇒x^2 (y^3 +1)=y^2 (x^3 +1) ⇒x^2 y^3 +x^2 = x^3 y^2 +y^2 ⇒x^2 y^2 (y−x)=y^2 −x^2 ⇒x^2 y^2 (y−x)−(y−x)(y+x)=0 ⇒(y−x)(x^2 y^2 −y−x)=0 for { ((y=x)),((x+y=1)) :}⇒x=y=(1/2) f=((9/8))^2 = ((81)/(64)) for { ((y+x=x^2 y^2 )),((y+x=1)) :}⇒x^2 y^2 =1 ⇒(1−x)^2 x^2 =1 ⇒(1−2x+x^2 )x^2 −1=0 ⇒x^4 −2x^3 +x^2 −1=0 ⇒ { ((x_1 =((1+(√5))/2)⇒y_1 =((1−(√5))/2))),((x_2 =((1−(√5))/2)⇒y_2 =((1+(√5))/2))) :} f={(((1+(√5))/2))^3 +1}{(((1−(√5))/2))^3 +1} f= 4 (maximal)
$$\:{f}\left({x},{y},\lambda\right)=\:\left({x}^{\mathrm{3}} +\mathrm{1}\right)\left({y}^{\mathrm{3}} +\mathrm{1}\right)+\lambda\left({x}+{y}−\mathrm{1}\right) \\ $$$$\:\frac{\partial{f}}{\partial{x}}\:=\:\mathrm{3}{x}^{\mathrm{2}} \left({y}^{\mathrm{3}} +\mathrm{1}\right)+\lambda=\mathrm{0} \\ $$$$\:\frac{\partial{f}}{\partial{y}}\:=\:\mathrm{3}{y}^{\mathrm{2}} \left({x}^{\mathrm{3}} +\mathrm{1}\right)+\lambda=\mathrm{0} \\ $$$$\:\frac{\partial{f}}{\partial\lambda}\:=\:{x}+{y}=\mathrm{1} \\ $$$$\Rightarrow\mathrm{3}{x}^{\mathrm{2}} \left({y}^{\mathrm{3}} +\mathrm{1}\right)=\mathrm{3}{y}^{\mathrm{2}} \left({x}^{\mathrm{3}} +\mathrm{1}\right) \\ $$$$\Rightarrow{x}^{\mathrm{2}} \left({y}^{\mathrm{3}} +\mathrm{1}\right)={y}^{\mathrm{2}} \left({x}^{\mathrm{3}} +\mathrm{1}\right) \\ $$$$\Rightarrow{x}^{\mathrm{2}} {y}^{\mathrm{3}} +{x}^{\mathrm{2}} =\:{x}^{\mathrm{3}} {y}^{\mathrm{2}} +{y}^{\mathrm{2}} \\ $$$$\Rightarrow{x}^{\mathrm{2}} {y}^{\mathrm{2}} \left({y}−{x}\right)={y}^{\mathrm{2}} −{x}^{\mathrm{2}} \\ $$$$\Rightarrow{x}^{\mathrm{2}} {y}^{\mathrm{2}} \left({y}−{x}\right)−\left({y}−{x}\right)\left({y}+{x}\right)=\mathrm{0} \\ $$$$\Rightarrow\left({y}−{x}\right)\left({x}^{\mathrm{2}} {y}^{\mathrm{2}} −{y}−{x}\right)=\mathrm{0} \\ $$$${for}\:\begin{cases}{{y}={x}}\\{{x}+{y}=\mathrm{1}}\end{cases}\Rightarrow{x}={y}=\frac{\mathrm{1}}{\mathrm{2}} \\ $$$$\:{f}=\left(\frac{\mathrm{9}}{\mathrm{8}}\right)^{\mathrm{2}} =\:\frac{\mathrm{81}}{\mathrm{64}} \\ $$$${for}\:\begin{cases}{{y}+{x}={x}^{\mathrm{2}} {y}^{\mathrm{2}} }\\{{y}+{x}=\mathrm{1}}\end{cases}\Rightarrow{x}^{\mathrm{2}} {y}^{\mathrm{2}} =\mathrm{1} \\ $$$$\Rightarrow\left(\mathrm{1}−{x}\right)^{\mathrm{2}} \:{x}^{\mathrm{2}} =\mathrm{1} \\ $$$$\Rightarrow\left(\mathrm{1}−\mathrm{2}{x}+{x}^{\mathrm{2}} \right){x}^{\mathrm{2}} −\mathrm{1}=\mathrm{0} \\ $$$$\Rightarrow{x}^{\mathrm{4}} −\mathrm{2}{x}^{\mathrm{3}} +{x}^{\mathrm{2}} −\mathrm{1}=\mathrm{0} \\ $$$$\Rightarrow\begin{cases}{{x}_{\mathrm{1}} =\frac{\mathrm{1}+\sqrt{\mathrm{5}}}{\mathrm{2}}\Rightarrow{y}_{\mathrm{1}} =\frac{\mathrm{1}−\sqrt{\mathrm{5}}}{\mathrm{2}}}\\{{x}_{\mathrm{2}} =\frac{\mathrm{1}−\sqrt{\mathrm{5}}}{\mathrm{2}}\Rightarrow{y}_{\mathrm{2}} =\frac{\mathrm{1}+\sqrt{\mathrm{5}}}{\mathrm{2}}}\end{cases} \\ $$$$\:{f}=\left\{\left(\frac{\mathrm{1}+\sqrt{\mathrm{5}}}{\mathrm{2}}\right)^{\mathrm{3}} +\mathrm{1}\right\}\left\{\left(\frac{\mathrm{1}−\sqrt{\mathrm{5}}}{\mathrm{2}}\right)^{\mathrm{3}} +\mathrm{1}\right\}\: \\ $$$$\:{f}=\:\mathrm{4}\:\left({maximal}\right)\: \\ $$$$ \\ $$
Commented by infinityaction last updated on 24/Apr/22
thank you sir
$${thank}\:{you}\:{sir} \\ $$
Commented by tahish last updated on 24/Apr/22
sir muje new image updolad krani ho to kaiser kare
Commented by infinityaction last updated on 26/Apr/22
meri talegram id @infinity_action hai aap vaha pr pucche
$${meri}\:{talegram}\:{id}\:@{infinity\_action} \\ $$$${hai}\:{aap}\:{vaha}\:{pr}\:{pucche} \\ $$
Answered by MJS_new last updated on 24/Apr/22
y=1−x ⇒ f(x)=(x^3 +1)×(y^3 +1)=(x^3 +1)×((1−x)^3 +1)= =(x^2 −x+1)(x+1)×(x^2 −x+1)(2−x)= =−(x−2)(x+1)(x^2 −x+1)^2 f′(x)=0 −3(2x−1)(x^2 −x−1)(x^2 −x+1)=0 x_1 =(1/2)−((√5)/2); f(x_1 )=4 x_2 =(1/2); f(x_2 )=((81)/(64)) x_3 =(1/2)+((√5)/2); f(x_3 )=4 max value is 4
$${y}=\mathrm{1}−{x} \\ $$$$\Rightarrow \\ $$$${f}\left({x}\right)=\left({x}^{\mathrm{3}} +\mathrm{1}\right)×\left({y}^{\mathrm{3}} +\mathrm{1}\right)=\left({x}^{\mathrm{3}} +\mathrm{1}\right)×\left(\left(\mathrm{1}−{x}\right)^{\mathrm{3}} +\mathrm{1}\right)= \\ $$$$=\left({x}^{\mathrm{2}} −{x}+\mathrm{1}\right)\left({x}+\mathrm{1}\right)×\left({x}^{\mathrm{2}} −{x}+\mathrm{1}\right)\left(\mathrm{2}−{x}\right)= \\ $$$$=−\left({x}−\mathrm{2}\right)\left({x}+\mathrm{1}\right)\left({x}^{\mathrm{2}} −{x}+\mathrm{1}\right)^{\mathrm{2}} \\ $$$$ \\ $$$${f}'\left({x}\right)=\mathrm{0} \\ $$$$−\mathrm{3}\left(\mathrm{2}{x}−\mathrm{1}\right)\left({x}^{\mathrm{2}} −{x}−\mathrm{1}\right)\left({x}^{\mathrm{2}} −{x}+\mathrm{1}\right)=\mathrm{0} \\ $$$${x}_{\mathrm{1}} =\frac{\mathrm{1}}{\mathrm{2}}−\frac{\sqrt{\mathrm{5}}}{\mathrm{2}};\:{f}\left({x}_{\mathrm{1}} \right)=\mathrm{4} \\ $$$${x}_{\mathrm{2}} =\frac{\mathrm{1}}{\mathrm{2}};\:{f}\left({x}_{\mathrm{2}} \right)=\frac{\mathrm{81}}{\mathrm{64}} \\ $$$${x}_{\mathrm{3}} =\frac{\mathrm{1}}{\mathrm{2}}+\frac{\sqrt{\mathrm{5}}}{\mathrm{2}};\:{f}\left({x}_{\mathrm{3}} \right)=\mathrm{4} \\ $$$$\mathrm{max}\:\mathrm{value}\:\mathrm{is}\:\mathrm{4} \\ $$
Commented by infinityaction last updated on 24/Apr/22
thank you sir
$${thank}\:{you}\:{sir} \\ $$
Answered by mr W last updated on 24/Apr/22
S=(x^3 +1)(y^3 +1) =1+(xy)^3 +x^3 +y^3 =1+(xy)^3 +(x+y)^3 −3xy(x+y) =1+(xy)^3 +1−3xy =(xy)^3 −3(xy)+2 (xy)^3 −3(xy)+2−S=0 Δ=(−1)^3 +(1−(S/2))^2 ≤0 −1≤1−(S/2)≤1 0≤S≤4 ⇒S_(max) =4
$${S}=\left({x}^{\mathrm{3}} +\mathrm{1}\right)\left({y}^{\mathrm{3}} +\mathrm{1}\right) \\ $$$$\:\:=\mathrm{1}+\left({xy}\right)^{\mathrm{3}} +{x}^{\mathrm{3}} +{y}^{\mathrm{3}} \\ $$$$\:\:=\mathrm{1}+\left({xy}\right)^{\mathrm{3}} +\left({x}+{y}\right)^{\mathrm{3}} −\mathrm{3}{xy}\left({x}+{y}\right) \\ $$$$\:\:=\mathrm{1}+\left({xy}\right)^{\mathrm{3}} +\mathrm{1}−\mathrm{3}{xy} \\ $$$$\:\:=\left({xy}\right)^{\mathrm{3}} −\mathrm{3}\left({xy}\right)+\mathrm{2} \\ $$$$\left({xy}\right)^{\mathrm{3}} −\mathrm{3}\left({xy}\right)+\mathrm{2}−{S}=\mathrm{0} \\ $$$$\Delta=\left(−\mathrm{1}\right)^{\mathrm{3}} +\left(\mathrm{1}−\frac{{S}}{\mathrm{2}}\right)^{\mathrm{2}} \leqslant\mathrm{0} \\ $$$$−\mathrm{1}\leqslant\mathrm{1}−\frac{{S}}{\mathrm{2}}\leqslant\mathrm{1} \\ $$$$\mathrm{0}\leqslant{S}\leqslant\mathrm{4}\:\Rightarrow{S}_{{max}} =\mathrm{4} \\ $$
Commented by infinityaction last updated on 25/Apr/22
thank you sir
$${thank}\:{you}\:{sir} \\ $$

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