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Question Number 169117 by infinityaction last updated on 24/Apr/22

	Answered by greougoury555 last updated on 24/Apr/22
	$f(x,y,λ)= (x^3 +1)(y^3 +1)+λ(x+y−1) (∂f/∂x) = 3x^2 (y^3 +1)+λ=0 (∂f/∂y) = 3y^2 (x^3 +1)+λ=0 (∂f/∂λ) = x+y=1 ⇒3x^2 (y^3 +1)=3y^2 (x^3 +1) ⇒x^2 (y^3 +1)=y^2 (x^3 +1) ⇒x^2 y^3 +x^2 = x^3 y^2 +y^2 ⇒x^2 y^2 (y−x)=y^2 −x^2 ⇒x^2 y^2 (y−x)−(y−x)(y+x)=0 ⇒(y−x)(x^2 y^2 −y−x)=0 for { ((y=x)),((x+y=1)) :}⇒x=y=(1/2) f=((9/8))^2 = ((81)/(64)) for { ((y+x=x^2 y^2 )),((y+x=1)) :}⇒x^2 y^2 =1 ⇒(1−x)^2 x^2 =1 ⇒(1−2x+x^2 )x^2 −1=0 ⇒x^4 −2x^3 +x^2 −1=0 ⇒ { ((x_1 =((1+(√5))/2)⇒y_1 =((1−(√5))/2))),((x_2 =((1−(√5))/2)⇒y_2 =((1+(√5))/2))) :} f={(((1+(√5))/2))^3 +1}{(((1−(√5))/2))^3 +1} f= 4 (maximal)$
	$f (x, y, λ) = (x^{3} + 1) (y^{3} + 1) + λ (x + y - 1)$ $\frac{\partial f}{\partial x} = 3 x^{2} (y^{3} + 1) + λ = 0$ $\frac{\partial f}{\partial y} = 3 y^{2} (x^{3} + 1) + λ = 0$ $\frac{\partial f}{\partial λ} = x + y = 1$ $\Rightarrow 3 x^{2} (y^{3} + 1) = 3 y^{2} (x^{3} + 1)$ $\Rightarrow x^{2} (y^{3} + 1) = y^{2} (x^{3} + 1)$ $\Rightarrow x^{2} y^{3} + x^{2} = x^{3} y^{2} + y^{2}$ $\Rightarrow x^{2} y^{2} (y - x) = y^{2} - x^{2}$ $\Rightarrow x^{2} y^{2} (y - x) - (y - x) (y + x) = 0$ $\Rightarrow (y - x) (x^{2} y^{2} - y - x) = 0$ $f o r {\begin{cases} y = x \\ x + y = 1 \end{cases} \Rightarrow x = y = \frac{1}{2}$ $f = {(\frac{9}{8})}^{2} = \frac{81}{64}$ $f o r {\begin{cases} y + x = x^{2} y^{2} \\ y + x = 1 \end{cases} \Rightarrow x^{2} y^{2} = 1$ $\Rightarrow {(1 - x)}^{2} x^{2} = 1$ $\Rightarrow (1 - 2 x + x^{2}) x^{2} - 1 = 0$ $\Rightarrow x^{4} - 2 x^{3} + x^{2} - 1 = 0$ $\Rightarrow {\begin{cases} x_{1} = \frac{1 + \sqrt{5}}{2} \Rightarrow y_{1} = \frac{1 - \sqrt{5}}{2} \\ x_{2} = \frac{1 - \sqrt{5}}{2} \Rightarrow y_{2} = \frac{1 + \sqrt{5}}{2} \end{cases}$ $f = {{(\frac{1 + \sqrt{5}}{2})}^{3} + 1} {{(\frac{1 - \sqrt{5}}{2})}^{3} + 1}$ $f = 4 (m a x i m a l)$
	Commented by infinityaction last updated on 24/Apr/22

	$t h a n k y o u s i r$
	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

	$m e r i t a l e g r a m i d @ i n f i n i t y_a c t i o n$ $h a i a a p v a h a p r p u c c h e$
	Answered by MJS_new last updated on 24/Apr/22

	$y = 1 - x$ $\Rightarrow$ $f (x) = (x^{3} + 1) \times (y^{3} + 1) = (x^{3} + 1) \times ({(1 - x)}^{3} + 1) =$ $= (x^{2} - x + 1) (x + 1) \times (x^{2} - x + 1) (2 - x) =$ $= - (x - 2) (x + 1) {(x^{2} - x + 1)}^{2}$ $f^{'} (x) = 0$ $- 3 (2 x - 1) (x^{2} - x - 1) (x^{2} - x + 1) = 0$ $x_{1} = \frac{1}{2} - \frac{\sqrt{5}}{2}; f (x_{1}) = 4$ $x_{2} = \frac{1}{2}; f (x_{2}) = \frac{81}{64}$ $x_{3} = \frac{1}{2} + \frac{\sqrt{5}}{2}; f (x_{3}) = 4$ $\max value is 4$
	Commented by infinityaction last updated on 24/Apr/22

	$t h a n k y o u s i r$
	Answered by mr W last updated on 24/Apr/22

	$S = (x^{3} + 1) (y^{3} + 1)$ $= 1 + {(x y)}^{3} + x^{3} + y^{3}$ $= 1 + {(x y)}^{3} + {(x + y)}^{3} - 3 x y (x + y)$ $= 1 + {(x y)}^{3} + 1 - 3 x y$ $= {(x y)}^{3} - 3 (x y) + 2$ ${(x y)}^{3} - 3 (x y) + 2 - S = 0$ $Δ = {(- 1)}^{3} + {(1 - \frac{S}{2})}^{2} ⩽ 0$ $- 1 ⩽ 1 - \frac{S}{2} ⩽ 1$ $0 ⩽ S ⩽ 4 \Rightarrow S_{m a x} = 4$
	Commented by infinityaction last updated on 25/Apr/22

	$t h a n k y o u s i r$
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