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Question Number 201613 by hardmath last updated on 09/Dec/23
x,y,z ∈ R { ((xy + yz + zx = 3)),((x + y + z = 5)) :} → max(z) = ?
$$\mathrm{x},\mathrm{y},\mathrm{z}\:\in\:\mathbb{R} \\ $$$$\begin{cases}{\mathrm{xy}\:+\:\mathrm{yz}\:+\:\mathrm{zx}\:=\:\mathrm{3}}\\{\mathrm{x}\:+\:\mathrm{y}\:+\:\mathrm{z}\:=\:\mathrm{5}}\end{cases}\:\:\:\:\:\rightarrow\:\:\:\:\mathrm{max}\left(\boldsymbol{\mathrm{z}}\right)\:=\:? \\ $$
Answered by aleks041103 last updated on 09/Dec/23
x+y+z=5⇒z=5−x−y ⇒xy+(x+y)(5−(x+y))=3 xy+5x+5y−x^2 −2xy−y^2 =3 ⇒x^2 +xy+y^2 −5x−5y+3=0 (1) max z ⇔ dz=0=−dx−dy⇒dx=−dy d (1) 2xdx+ydx+xdy+2ydy−5dx−5dy=0 (2x+y−5)dx+(x+2y−5)dy=0 dx=−dy ⇒(2x+y−5)−(x+2y−5)=x−y=0 ⇒x=y (1)⇒x^2 +x^2 +x^2 −5x−5x+3=0 3x^2 −10x+3=0 x_(1,2) =((10±(√(100−4.3.3)))/6)=((5±4)/3)=(1/3);3 ⇒z_1 =5−2x_1 =((13)/3) z_2 =5−2x_2 =−1 ⇒max(z)=((13)/3)
$${x}+{y}+{z}=\mathrm{5}\Rightarrow{z}=\mathrm{5}−{x}−{y} \\ $$$$\Rightarrow{xy}+\left({x}+{y}\right)\left(\mathrm{5}−\left({x}+{y}\right)\right)=\mathrm{3} \\ $$$${xy}+\mathrm{5}{x}+\mathrm{5}{y}−{x}^{\mathrm{2}} −\mathrm{2}{xy}−{y}^{\mathrm{2}} =\mathrm{3} \\ $$$$\Rightarrow{x}^{\mathrm{2}} +{xy}+{y}^{\mathrm{2}} −\mathrm{5}{x}−\mathrm{5}{y}+\mathrm{3}=\mathrm{0}\:\left(\mathrm{1}\right) \\ $$$${max}\:{z}\:\Leftrightarrow\:{dz}=\mathrm{0}=−{dx}−{dy}\Rightarrow{dx}=−{dy} \\ $$$${d}\:\left(\mathrm{1}\right) \\ $$$$\mathrm{2}{xdx}+{ydx}+{xdy}+\mathrm{2}{ydy}−\mathrm{5}{dx}−\mathrm{5}{dy}=\mathrm{0} \\ $$$$\left(\mathrm{2}{x}+{y}−\mathrm{5}\right){dx}+\left({x}+\mathrm{2}{y}−\mathrm{5}\right){dy}=\mathrm{0} \\ $$$${dx}=−{dy} \\ $$$$\Rightarrow\left(\mathrm{2}{x}+{y}−\mathrm{5}\right)−\left({x}+\mathrm{2}{y}−\mathrm{5}\right)={x}−{y}=\mathrm{0} \\ $$$$\Rightarrow{x}={y} \\ $$$$\left(\mathrm{1}\right)\Rightarrow{x}^{\mathrm{2}} +{x}^{\mathrm{2}} +{x}^{\mathrm{2}} −\mathrm{5}{x}−\mathrm{5}{x}+\mathrm{3}=\mathrm{0} \\ $$$$\mathrm{3}{x}^{\mathrm{2}} −\mathrm{10}{x}+\mathrm{3}=\mathrm{0} \\ $$$${x}_{\mathrm{1},\mathrm{2}} =\frac{\mathrm{10}\pm\sqrt{\mathrm{100}−\mathrm{4}.\mathrm{3}.\mathrm{3}}}{\mathrm{6}}=\frac{\mathrm{5}\pm\mathrm{4}}{\mathrm{3}}=\frac{\mathrm{1}}{\mathrm{3}};\mathrm{3} \\ $$$$\Rightarrow{z}_{\mathrm{1}} =\mathrm{5}−\mathrm{2}{x}_{\mathrm{1}} =\frac{\mathrm{13}}{\mathrm{3}} \\ $$$${z}_{\mathrm{2}} =\mathrm{5}−\mathrm{2}{x}_{\mathrm{2}} =−\mathrm{1} \\ $$$$\Rightarrow{max}\left({z}\right)=\frac{\mathrm{13}}{\mathrm{3}} \\ $$
Commented by aleks041103 last updated on 09/Dec/23
the geometric idea is: (1) → ellipse On another note: z=5−(x+y) x+y= ((x),(y) ) . ((1),(1) ) =(√2) n^→ .r^→ ⇒x+y=(√2)×(distance from the line x+y=0) ⇒z=5−(x+y) is max(min) when the distance of (x,y) from the line x+y=0 is min(max). ⇒we want to find the points at which the tangents to the ellipse are parallel to the ellipse.
$${the}\:{geometric}\:{idea}\:{is}: \\ $$$$\left(\mathrm{1}\right)\:\rightarrow\:{ellipse} \\ $$$${On}\:{another}\:{note}: \\ $$$${z}=\mathrm{5}−\left({x}+{y}\right) \\ $$$${x}+{y}=\begin{pmatrix}{{x}}\\{{y}}\end{pmatrix}\:.\:\begin{pmatrix}{\mathrm{1}}\\{\mathrm{1}}\end{pmatrix}\:=\sqrt{\mathrm{2}}\:\overset{\rightarrow} {{n}}.\overset{\rightarrow} {{r}} \\ $$$$\Rightarrow{x}+{y}=\sqrt{\mathrm{2}}×\left({distance}\:{from}\:{the}\:{line}\:{x}+{y}=\mathrm{0}\right) \\ $$$$\Rightarrow{z}=\mathrm{5}−\left({x}+{y}\right)\:{is}\:{max}\left({min}\right) \\ $$$${when}\:{the}\:{distance}\:{of}\:\left({x},{y}\right)\:{from}\:{the}\:{line} \\ $$$${x}+{y}=\mathrm{0}\:{is}\:{min}\left({max}\right). \\ $$$$\Rightarrow{we}\:{want}\:{to}\:{find}\:{the}\:{points} \\ $$$${at}\:{which}\:{the}\:{tangents}\:{to}\:{the}\:{ellipse} \\ $$$${are}\:{parallel}\:{to}\:{the}\:{ellipse}. \\ $$
Commented by aleks041103 last updated on 09/Dec/23
Commented by hardmath last updated on 10/Dec/23
perfect dear professor thank you so much
$$\mathrm{perfect}\:\mathrm{dear}\:\mathrm{professor}\:\mathrm{thank}\:\mathrm{you}\:\mathrm{so}\:\mathrm{much} \\ $$
Answered by mr W last updated on 10/Dec/23
say t=x+y z=5−t to find max(z), we only need to find min(t). xy+(x+y)z=3 xy+(x+y)(5−(x+y))=3 (x+y)^2 −5(x+y)+3=xy≤(((x+y)/2))^2 =(((x+y)^2 )/4) 3(x+y)^2 −20(x+y)+12≤0 or 3t^2 −20t+12≤0 (3t−2)(t−6)≤0 ⇒(2/3)≤t≤6 i.e. min(t)=(2/3), max(t)=6 ⇒max(z)=5−(2/3)=((13)/3) ⇒min(z)=5−6=−1
$${say}\:{t}={x}+{y} \\ $$$${z}=\mathrm{5}−{t} \\ $$$${to}\:{find}\:{max}\left({z}\right),\:{we}\:{only}\:{need}\:{to}\:{find} \\ $$$${min}\left({t}\right). \\ $$$${xy}+\left({x}+{y}\right){z}=\mathrm{3} \\ $$$${xy}+\left({x}+{y}\right)\left(\mathrm{5}−\left({x}+{y}\right)\right)=\mathrm{3} \\ $$$$\left({x}+{y}\right)^{\mathrm{2}} −\mathrm{5}\left({x}+{y}\right)+\mathrm{3}={xy}\leqslant\left(\frac{{x}+{y}}{\mathrm{2}}\right)^{\mathrm{2}} =\frac{\left({x}+{y}\right)^{\mathrm{2}} }{\mathrm{4}} \\ $$$$\mathrm{3}\left({x}+{y}\right)^{\mathrm{2}} −\mathrm{20}\left({x}+{y}\right)+\mathrm{12}\leqslant\mathrm{0} \\ $$$${or}\:\mathrm{3}{t}^{\mathrm{2}} −\mathrm{20}{t}+\mathrm{12}\leqslant\mathrm{0} \\ $$$$\left(\mathrm{3}{t}−\mathrm{2}\right)\left({t}−\mathrm{6}\right)\leqslant\mathrm{0} \\ $$$$\Rightarrow\frac{\mathrm{2}}{\mathrm{3}}\leqslant{t}\leqslant\mathrm{6} \\ $$$${i}.{e}.\:{min}\left({t}\right)=\frac{\mathrm{2}}{\mathrm{3}},\:{max}\left({t}\right)=\mathrm{6} \\ $$$$\Rightarrow{max}\left({z}\right)=\mathrm{5}−\frac{\mathrm{2}}{\mathrm{3}}=\frac{\mathrm{13}}{\mathrm{3}} \\ $$$$\Rightarrow{min}\left({z}\right)=\mathrm{5}−\mathrm{6}=−\mathrm{1} \\ $$
Commented by mr W last updated on 10/Dec/23
A.M.≥G.M. ((a+b)/2)≥(√(ab)) ⇒ab≤(((a+b)/2))^2
$${A}.{M}.\geqslant{G}.{M}. \\ $$$$\frac{{a}+{b}}{\mathrm{2}}\geqslant\sqrt{{ab}}\:\Rightarrow{ab}\leqslant\left(\frac{{a}+{b}}{\mathrm{2}}\right)^{\mathrm{2}} \\ $$
Commented by hardmath last updated on 10/Dec/23
perfect dear professor thank you so much
$$\mathrm{perfect}\:\mathrm{dear}\:\mathrm{professor}\:\mathrm{thank}\:\mathrm{you}\:\mathrm{so}\:\mathrm{much} \\ $$

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