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Question Number 190546 by cortano12 last updated on 05/Apr/23
Given x,y,z>0 and x^2 +y^2 +z^2 +x+2y+3z=23 find maximum of x+y+z.
$$\:\mathrm{Given}\:\mathrm{x},\mathrm{y},\mathrm{z}>\mathrm{0}\:\mathrm{and}\: \\ $$$$\:\mathrm{x}^{\mathrm{2}} +\mathrm{y}^{\mathrm{2}} +\mathrm{z}^{\mathrm{2}} +\mathrm{x}+\mathrm{2y}+\mathrm{3z}=\mathrm{23}\: \\ $$$$\:\mathrm{find}\:\mathrm{maximum}\:\mathrm{of}\:\mathrm{x}+\mathrm{y}+\mathrm{z}. \\ $$
Answered by mr W last updated on 05/Apr/23
(x+(1/2))^2 +(y+1)^2 +(z+(3/2))^2 =((53)/2) say x+y+z=s (√((53)/2))=((∣−(1/2)−1−(3/2)−s∣)/( (√3))) s_(max) =−3+(√((159)/2)) ✓ s_(min) =−3−(√((159)/2))
$$\left({x}+\frac{\mathrm{1}}{\mathrm{2}}\right)^{\mathrm{2}} +\left({y}+\mathrm{1}\right)^{\mathrm{2}} +\left({z}+\frac{\mathrm{3}}{\mathrm{2}}\right)^{\mathrm{2}} =\frac{\mathrm{53}}{\mathrm{2}} \\ $$$${say}\:{x}+{y}+{z}={s} \\ $$$$\sqrt{\frac{\mathrm{53}}{\mathrm{2}}}=\frac{\mid−\frac{\mathrm{1}}{\mathrm{2}}−\mathrm{1}−\frac{\mathrm{3}}{\mathrm{2}}−{s}\mid}{\:\sqrt{\mathrm{3}}} \\ $$$${s}_{{max}} =−\mathrm{3}+\sqrt{\frac{\mathrm{159}}{\mathrm{2}}}\:\checkmark \\ $$$${s}_{{min}} =−\mathrm{3}−\sqrt{\frac{\mathrm{159}}{\mathrm{2}}} \\ $$
Commented by cortano12 last updated on 05/Apr/23
by Cauhcy?
$$\mathrm{by}\:\mathrm{Cauhcy}? \\ $$
Commented by mr W last updated on 06/Apr/23
no. plane x+y+z=s must tangent the sphere (x+(1/2))^2 +(y+1)^2 +(z+(3/2))^2 =((53)/2) ⇒distance from center of sphere (−(1/2),−1,−(3/2)) to the plane is the radius (√((53)/2)).
$${no}. \\ $$$${plane}\:{x}+{y}+{z}={s}\:{must}\:{tangent}\:{the} \\ $$$${sphere}\:\left({x}+\frac{\mathrm{1}}{\mathrm{2}}\right)^{\mathrm{2}} +\left({y}+\mathrm{1}\right)^{\mathrm{2}} +\left({z}+\frac{\mathrm{3}}{\mathrm{2}}\right)^{\mathrm{2}} =\frac{\mathrm{53}}{\mathrm{2}} \\ $$$$\Rightarrow{distance}\:{from}\:{center}\:{of}\:{sphere} \\ $$$$\left(−\frac{\mathrm{1}}{\mathrm{2}},−\mathrm{1},−\frac{\mathrm{3}}{\mathrm{2}}\right)\:{to}\:{the}\:{plane}\:{is}\:{the} \\ $$$${radius}\:\sqrt{\frac{\mathrm{53}}{\mathrm{2}}}. \\ $$
Commented by cortano12 last updated on 07/Apr/23
ooyes.thank you
$$\mathrm{ooyes}.\mathrm{thank}\:\mathrm{you} \\ $$
Answered by mr W last updated on 07/Apr/23
Method II F=x+y+z−(1/λ)(x^2 +y^2 +z^2 +x+2y+3z−23) (∂F/∂x)=1−(1/λ)(2x+1)=0 ⇒x=(1/2)(λ−1) (∂F/∂y)=1−(1/λ)(2y+2)=0 ⇒y=(1/2)(λ−2) (∂F/∂z)=1−(1/λ)(2z+3)=0 ⇒z=(1/2)(λ−3) (((λ−1)^2 )/4)+(((λ−1))/2)+(((λ−2)^2 )/4)+((2(λ−2))/2)+(((λ−3)^2 )/4)+((3(λ−3))/2)=23 3λ^2 =106 ⇒λ=±(√((106)/3)) x+y+z=−3+((3λ)/2)=−3±(√((159)/2)) (x+y+z)_(max) =−3+(√((159)/2)) (x+y+z)_(min) =−3−(√((159)/2))
$${Method}\:{II} \\ $$$${F}={x}+{y}+{z}−\frac{\mathrm{1}}{\lambda}\left({x}^{\mathrm{2}} +{y}^{\mathrm{2}} +{z}^{\mathrm{2}} +{x}+\mathrm{2}{y}+\mathrm{3}{z}−\mathrm{23}\right) \\ $$$$\frac{\partial{F}}{\partial{x}}=\mathrm{1}−\frac{\mathrm{1}}{\lambda}\left(\mathrm{2}{x}+\mathrm{1}\right)=\mathrm{0}\:\Rightarrow{x}=\frac{\mathrm{1}}{\mathrm{2}}\left(\lambda−\mathrm{1}\right) \\ $$$$\frac{\partial{F}}{\partial{y}}=\mathrm{1}−\frac{\mathrm{1}}{\lambda}\left(\mathrm{2}{y}+\mathrm{2}\right)=\mathrm{0}\:\Rightarrow{y}=\frac{\mathrm{1}}{\mathrm{2}}\left(\lambda−\mathrm{2}\right) \\ $$$$\frac{\partial{F}}{\partial{z}}=\mathrm{1}−\frac{\mathrm{1}}{\lambda}\left(\mathrm{2}{z}+\mathrm{3}\right)=\mathrm{0}\:\Rightarrow{z}=\frac{\mathrm{1}}{\mathrm{2}}\left(\lambda−\mathrm{3}\right) \\ $$$$\frac{\left(\lambda−\mathrm{1}\right)^{\mathrm{2}} }{\mathrm{4}}+\frac{\left(\lambda−\mathrm{1}\right)}{\mathrm{2}}+\frac{\left(\lambda−\mathrm{2}\right)^{\mathrm{2}} }{\mathrm{4}}+\frac{\mathrm{2}\left(\lambda−\mathrm{2}\right)}{\mathrm{2}}+\frac{\left(\lambda−\mathrm{3}\right)^{\mathrm{2}} }{\mathrm{4}}+\frac{\mathrm{3}\left(\lambda−\mathrm{3}\right)}{\mathrm{2}}=\mathrm{23} \\ $$$$\mathrm{3}\lambda^{\mathrm{2}} =\mathrm{106} \\ $$$$\Rightarrow\lambda=\pm\sqrt{\frac{\mathrm{106}}{\mathrm{3}}} \\ $$$${x}+{y}+{z}=−\mathrm{3}+\frac{\mathrm{3}\lambda}{\mathrm{2}}=−\mathrm{3}\pm\sqrt{\frac{\mathrm{159}}{\mathrm{2}}} \\ $$$$\left({x}+{y}+{z}\right)_{{max}} =−\mathrm{3}+\sqrt{\frac{\mathrm{159}}{\mathrm{2}}} \\ $$$$\left({x}+{y}+{z}\right)_{{min}} =−\mathrm{3}−\sqrt{\frac{\mathrm{159}}{\mathrm{2}}} \\ $$

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