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Question Number 208553 by efronzo1 last updated on 18/Jun/24

Answered by Frix last updated on 18/Jun/24

x=u−v∧y=u+v  z=2(u^2 −v^2 +u)  3u^2 +v^2 =1 ⇒ v^2 =1−3u^2   z=2(4u^2 +u−1)  v∈R ⇒ −((√3)/3)≤u≤((√3)/3)  ⇒ ((2−2(√3))/3)≤z≤((2+2(√3))/3)  BUT  z′=16u+2=0 ⇒ u=−(1/8) ∧ z′′=16>0  ⇒ min(z)=−((17)/8)  ⇒ −((17)/8)≤f(x, y)≤((2+2(√3))/3)

x=uvy=u+vz=2(u2v2+u)3u2+v2=1v2=13u2z=2(4u2+u1)vR33u332233z2+233BUTz=16u+2=0u=18z=16>0min(z)=178178f(x,y)2+233

Commented by efronzo1 last updated on 18/Jun/24

yes min value is −((17)/8)

yesminvalueis178

Answered by A5T last updated on 18/Jun/24

Suppose y=f(x)  Then x^2 +y^2 +xy=1 ≡ x^2 +(f(x))^2 +xf(x)=1  (d/dx) of sides⇒2x+2f(x)f^′ (x)+xf^′ (x)+f(x)=0  ⇒f^′ (x)(2y+x)=−2x−y⇒f^′ (x)=((−2x−y)/(2y+x))...(i)    Similarly,x+2xy+y=x+2xf(x)+f(x)  (d/dx) of both sides⇒1+2xf′x+2f(x)+f^′ (x)=0  f^′ (x)=((−2y−1)/(2x+1))...(ii)  (i)=(ii)⇒f^′ (x)=((−2y−1)/(2x+1))=((−2x−y)/(2y+x))[x≠((−1)/2),2y≠−x]  ⇒4(x−y)(x+y)=−x+y=−(x−y)  ⇒y=x or 4(x+y)=−1  y=x⇒3x^2 =1⇒x=+_− ((√3)/3)=y⇒f(x,y)=((2+_− (√3))/3)  y=((−1)/4)−x⇒x=((+_− (√(61))−1)/8)⇒f(x,y)=((−17)/8)  Comparing values⇒max(f(x,y))=((2+(√3))/3)  and min(f(x,y))=((−17)/8)

Supposey=f(x)Thenx2+y2+xy=1x2+(f(x))2+xf(x)=1ddxofsides2x+2f(x)f(x)+xf(x)+f(x)=0f(x)(2y+x)=2xyf(x)=2xy2y+x...(i)Similarly,x+2xy+y=x+2xf(x)+f(x)ddxofbothsides1+2xfx+2f(x)+f(x)=0f(x)=2y12x+1...(ii)(i)=(ii)f(x)=2y12x+1=2xy2y+x[x12,2yx]4(xy)(x+y)=x+y=(xy)y=xor4(x+y)=1y=x3x2=1x=+33=yf(x,y)=2+33y=14xx=+6118f(x,y)=178Comparingvaluesmax(f(x,y))=2+33andmin(f(x,y))=178

Answered by efronzo1 last updated on 18/Jun/24

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