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Max-and-min-P-2-x-3-y-subject-to-constraint-x-2-9-y-2-25-1-x-2-y-2-

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Question Number 172253 by cortano1 last updated on 25/Jun/22
Max and min P=(√2) x+ (√3) y subject to constraint (x^2 /9) +(y^2 /(25)) ≤ 1 ≤ x^2 +y^2
$$\:\:{Max}\:{and}\:{min}\:{P}=\sqrt{\mathrm{2}}\:{x}+\:\sqrt{\mathrm{3}}\:{y} \\ $$$${subject}\:{to}\:{constraint}\: \\ $$$$\:\frac{{x}^{\mathrm{2}} }{\mathrm{9}}\:+\frac{{y}^{\mathrm{2}} }{\mathrm{25}}\:\leqslant\:\mathrm{1}\:\leqslant\:{x}^{\mathrm{2}} +{y}^{\mathrm{2}} \\ $$
Answered by mr W last updated on 25/Jun/22
Method I y=((P−(√2)x)/( (√3))) (x^2 /9)+(1/(25))(((P−(√2)x)/( (√3))))^2 =1 25x^2 +3(P^2 −2(√2)Px+2x^2 )=9×25 31x^2 −6(√2)Px+3P^2 −225=0 Δ=36×2P^2 −4×31(3P^2 −225)=0 P^2 =93 P=±(√(93)) ⇒P_(max) =(√(93)) ⇒P_(min) =−(√(93))
$${Method}\:{I} \\ $$$${y}=\frac{{P}−\sqrt{\mathrm{2}}{x}}{\:\sqrt{\mathrm{3}}} \\ $$$$\frac{{x}^{\mathrm{2}} }{\mathrm{9}}+\frac{\mathrm{1}}{\mathrm{25}}\left(\frac{{P}−\sqrt{\mathrm{2}}{x}}{\:\sqrt{\mathrm{3}}}\right)^{\mathrm{2}} =\mathrm{1} \\ $$$$\mathrm{25}{x}^{\mathrm{2}} +\mathrm{3}\left({P}^{\mathrm{2}} −\mathrm{2}\sqrt{\mathrm{2}}{Px}+\mathrm{2}{x}^{\mathrm{2}} \right)=\mathrm{9}×\mathrm{25} \\ $$$$\mathrm{31}{x}^{\mathrm{2}} −\mathrm{6}\sqrt{\mathrm{2}}{Px}+\mathrm{3}{P}^{\mathrm{2}} −\mathrm{225}=\mathrm{0} \\ $$$$\Delta=\mathrm{36}×\mathrm{2}{P}^{\mathrm{2}} −\mathrm{4}×\mathrm{31}\left(\mathrm{3}{P}^{\mathrm{2}} −\mathrm{225}\right)=\mathrm{0} \\ $$$${P}^{\mathrm{2}} =\mathrm{93} \\ $$$${P}=\pm\sqrt{\mathrm{93}} \\ $$$$\Rightarrow{P}_{{max}} =\sqrt{\mathrm{93}} \\ $$$$\Rightarrow{P}_{{min}} =−\sqrt{\mathrm{93}} \\ $$
Commented by mr W last updated on 25/Jun/22
Commented by Tawa11 last updated on 25/Jun/22
Great sir
$$\mathrm{Great}\:\mathrm{sir} \\ $$
Answered by mr W last updated on 25/Jun/22
Method II (x^2 /3^2 )+(y^2 /5^2 )=1 ⇒x=3 cos θ, y=5 sin θ P=(√2)x+(√3)y=3(√2) cos θ+5(√3) sin θ P=(√((3(√2))^2 +(5(√3))^2 ))(((3(√2))/( (√((3(√2))^2 +(5(√3))^2 ))))cos θ+((5(√3))/( (√((3(√2))^2 +(5(√3))^2 )))) sin θ) P=(√(93)) (((3(√2))/( (√(93 ))))cos θ+((5(√3))/( (√(93 )))) sin θ) P=(√(93)) sin (θ+tan^(−1) ((3(√2))/(5(√3)))) ⇒P_(max) =(√(93)) at θ=(π/2)−tan^(−1) ((3(√2))/(5(√3))) ⇒P_(min) =−(√(93)) at θ=−(π/2)−tan^(−1) ((3(√2))/(5(√3)))
$${Method}\:{II} \\ $$$$\frac{{x}^{\mathrm{2}} }{\mathrm{3}^{\mathrm{2}} }+\frac{{y}^{\mathrm{2}} }{\mathrm{5}^{\mathrm{2}} }=\mathrm{1} \\ $$$$\Rightarrow{x}=\mathrm{3}\:\mathrm{cos}\:\theta,\:{y}=\mathrm{5}\:\mathrm{sin}\:\theta \\ $$$${P}=\sqrt{\mathrm{2}}{x}+\sqrt{\mathrm{3}}{y}=\mathrm{3}\sqrt{\mathrm{2}}\:\mathrm{cos}\:\theta+\mathrm{5}\sqrt{\mathrm{3}}\:\mathrm{sin}\:\theta \\ $$$${P}=\sqrt{\left(\mathrm{3}\sqrt{\mathrm{2}}\right)^{\mathrm{2}} +\left(\mathrm{5}\sqrt{\mathrm{3}}\right)^{\mathrm{2}} \:}\left(\frac{\mathrm{3}\sqrt{\mathrm{2}}}{\:\sqrt{\left(\mathrm{3}\sqrt{\mathrm{2}}\right)^{\mathrm{2}} +\left(\mathrm{5}\sqrt{\mathrm{3}}\right)^{\mathrm{2}} \:}}\mathrm{cos}\:\theta+\frac{\mathrm{5}\sqrt{\mathrm{3}}}{\:\sqrt{\left(\mathrm{3}\sqrt{\mathrm{2}}\right)^{\mathrm{2}} +\left(\mathrm{5}\sqrt{\mathrm{3}}\right)^{\mathrm{2}} \:}}\:\mathrm{sin}\:\theta\right) \\ $$$${P}=\sqrt{\mathrm{93}}\:\left(\frac{\mathrm{3}\sqrt{\mathrm{2}}}{\:\sqrt{\mathrm{93}\:}}\mathrm{cos}\:\theta+\frac{\mathrm{5}\sqrt{\mathrm{3}}}{\:\sqrt{\mathrm{93}\:}}\:\mathrm{sin}\:\theta\right) \\ $$$${P}=\sqrt{\mathrm{93}}\:\mathrm{sin}\:\left(\theta+\mathrm{tan}^{−\mathrm{1}} \frac{\mathrm{3}\sqrt{\mathrm{2}}}{\mathrm{5}\sqrt{\mathrm{3}}}\right) \\ $$$$\Rightarrow{P}_{{max}} =\sqrt{\mathrm{93}}\:\:{at}\:\theta=\frac{\pi}{\mathrm{2}}−\mathrm{tan}^{−\mathrm{1}} \frac{\mathrm{3}\sqrt{\mathrm{2}}}{\mathrm{5}\sqrt{\mathrm{3}}} \\ $$$$\Rightarrow{P}_{{min}} =−\sqrt{\mathrm{93}}\:\:{at}\:\theta=−\frac{\pi}{\mathrm{2}}−\mathrm{tan}^{−\mathrm{1}} \frac{\mathrm{3}\sqrt{\mathrm{2}}}{\mathrm{5}\sqrt{\mathrm{3}}} \\ $$
Commented by mr W last updated on 25/Jun/22

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