Menu Close

Question-191651

Tinku Tara 0 Comments
Pin




Question Number 191651 by 073 last updated on 28/Apr/23
Commented by 073 last updated on 28/Apr/23
solution please???
$$\mathrm{solution}\:\mathrm{please}??? \\ $$
Answered by aleks041103 last updated on 28/Apr/23
That is to minimize g(x,y,z)=5x^2 +y^2 +2z^2 subject to x+y+z=34 We use Lagrange multipliers: F(x,y,z,λ)=g(x,y,z)+λ(x+y+z−34) and we want dF=0 (∂F/∂x)=10x+λ=0⇒x=−(λ/(10)) (∂F/∂y)=2y+λ=0⇒y=−(λ/2) (∂F/∂z)=4z+λ=0⇒z=−(λ/4) (∂F/∂λ)=x+y+z−34=0⇒λ((1/(10))+(1/2)+(1/4))+34=0 ⇒((17λ)/(20))=−34 ⇒λ=−40 ⇒(x,y,z)=(4,20,10) Since there is only one extrema, then it better be the one we want. ⇒f(x,y,z) is max at P=(x,y,z)=(4,20,10), where f(P)=900−5×16−400−200=220
$${That}\:{is}\:{to}\:{minimize}\:{g}\left({x},{y},{z}\right)=\mathrm{5}{x}^{\mathrm{2}} +{y}^{\mathrm{2}} +\mathrm{2}{z}^{\mathrm{2}} \\ $$$${subject}\:{to}\:{x}+{y}+{z}=\mathrm{34} \\ $$$${We}\:{use}\:{Lagrange}\:{multipliers}: \\ $$$${F}\left({x},{y},{z},\lambda\right)={g}\left({x},{y},{z}\right)+\lambda\left({x}+{y}+{z}−\mathrm{34}\right) \\ $$$${and}\:{we}\:{want}\:{dF}=\mathrm{0} \\ $$$$\frac{\partial{F}}{\partial{x}}=\mathrm{10}{x}+\lambda=\mathrm{0}\Rightarrow{x}=−\frac{\lambda}{\mathrm{10}} \\ $$$$\frac{\partial{F}}{\partial{y}}=\mathrm{2}{y}+\lambda=\mathrm{0}\Rightarrow{y}=−\frac{\lambda}{\mathrm{2}} \\ $$$$\frac{\partial{F}}{\partial{z}}=\mathrm{4}{z}+\lambda=\mathrm{0}\Rightarrow{z}=−\frac{\lambda}{\mathrm{4}} \\ $$$$\frac{\partial{F}}{\partial\lambda}={x}+{y}+{z}−\mathrm{34}=\mathrm{0}\Rightarrow\lambda\left(\frac{\mathrm{1}}{\mathrm{10}}+\frac{\mathrm{1}}{\mathrm{2}}+\frac{\mathrm{1}}{\mathrm{4}}\right)+\mathrm{34}=\mathrm{0} \\ $$$$\Rightarrow\frac{\mathrm{17}\lambda}{\mathrm{20}}=−\mathrm{34} \\ $$$$\Rightarrow\lambda=−\mathrm{40} \\ $$$$\Rightarrow\left({x},{y},{z}\right)=\left(\mathrm{4},\mathrm{20},\mathrm{10}\right) \\ $$$${Since}\:{there}\:{is}\:{only}\:{one}\:{extrema}, \\ $$$${then}\:{it}\:{better}\:{be}\:{the}\:{one}\:{we}\:{want}. \\ $$$$\Rightarrow{f}\left({x},{y},{z}\right)\:{is}\:{max}\:{at}\:{P}=\left({x},{y},{z}\right)=\left(\mathrm{4},\mathrm{20},\mathrm{10}\right), \\ $$$${where}\: \\ $$$${f}\left({P}\right)=\mathrm{900}−\mathrm{5}×\mathrm{16}−\mathrm{400}−\mathrm{200}=\mathrm{220} \\ $$$$ \\ $$
Answered by mr W last updated on 28/Apr/23
let p=(√5)x, q=y, r=(√2)z S=5x^2 +y^2 +2z^2 =p^2 +q^2 +r^2 x+y+z=34 ⇒(p/( (√5)))+q+(r/( (√2)))=34 ⇒(√2)p+(√(10))q+(√5)r−34(√(10))=0 (√S)=(√(p^2 +q^2 +r^2 )) is distance from (0,0,0) to (p,q,r). distance from (0,0,0) to plane (√2)p+(√(10))q+(√5)r−34(√(10))=0 is d=((∣(√2)×0+(√(10))×0+(√5)×0−34(√(10))∣)/( (√(((√2))^2 +((√(10)))^2 +((√5))^2 ))))=((34(√(10)))/( (√(17))))=2(√(170)) S_(min) =d^2 =680 f(x,y,z)=900−S f(x,y,z)_(max) =900−S_(min) =900−680=220 ✓
$${let}\:{p}=\sqrt{\mathrm{5}}{x},\:{q}={y},\:{r}=\sqrt{\mathrm{2}}{z} \\ $$$${S}=\mathrm{5}{x}^{\mathrm{2}} +{y}^{\mathrm{2}} +\mathrm{2}{z}^{\mathrm{2}} ={p}^{\mathrm{2}} +{q}^{\mathrm{2}} +{r}^{\mathrm{2}} \\ $$$${x}+{y}+{z}=\mathrm{34}\: \\ $$$$\Rightarrow\frac{{p}}{\:\sqrt{\mathrm{5}}}+{q}+\frac{{r}}{\:\sqrt{\mathrm{2}}}=\mathrm{34}\: \\ $$$$\Rightarrow\sqrt{\mathrm{2}}{p}+\sqrt{\mathrm{10}}{q}+\sqrt{\mathrm{5}}{r}−\mathrm{34}\sqrt{\mathrm{10}}=\mathrm{0} \\ $$$$ \\ $$$$\sqrt{{S}}=\sqrt{{p}^{\mathrm{2}} +{q}^{\mathrm{2}} +{r}^{\mathrm{2}} }\:{is}\:{distance}\:{from} \\ $$$$\left(\mathrm{0},\mathrm{0},\mathrm{0}\right)\:{to}\:\left({p},{q},{r}\right). \\ $$$${distance}\:{from}\:\left(\mathrm{0},\mathrm{0},\mathrm{0}\right)\:{to}\:{plane} \\ $$$$\sqrt{\mathrm{2}}{p}+\sqrt{\mathrm{10}}{q}+\sqrt{\mathrm{5}}{r}−\mathrm{34}\sqrt{\mathrm{10}}=\mathrm{0} \\ $$$${is}\:{d}=\frac{\mid\sqrt{\mathrm{2}}×\mathrm{0}+\sqrt{\mathrm{10}}×\mathrm{0}+\sqrt{\mathrm{5}}×\mathrm{0}−\mathrm{34}\sqrt{\mathrm{10}}\mid}{\:\sqrt{\left(\sqrt{\mathrm{2}}\right)^{\mathrm{2}} +\left(\sqrt{\mathrm{10}}\right)^{\mathrm{2}} +\left(\sqrt{\mathrm{5}}\right)^{\mathrm{2}} }}=\frac{\mathrm{34}\sqrt{\mathrm{10}}}{\:\sqrt{\mathrm{17}}}=\mathrm{2}\sqrt{\mathrm{170}} \\ $$$${S}_{{min}} ={d}^{\mathrm{2}} =\mathrm{680} \\ $$$${f}\left({x},{y},{z}\right)=\mathrm{900}−{S} \\ $$$${f}\left({x},{y},{z}\right)_{{max}} =\mathrm{900}−{S}_{{min}} =\mathrm{900}−\mathrm{680}=\mathrm{220}\:\checkmark \\ $$

Pin

Leave a Reply

Your email address will not be published. Required fields are marked *