Question and Answers Forum

All Questions      Topic List

Differentiation Questions

Previous in All Question      Next in All Question      

Previous in Differentiation      Next in Differentiation      

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

Terms of Service

Privacy Policy

Contact: info@tinkutara.com