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Question Number 148027 by mathdanisur last updated on 25/Jul/21
if 2x^3 + 3y^3 = 200 ⇒ (x^2 y^2 )_(max) = ?
$${if}\:\:\:\mathrm{2}{x}^{\mathrm{3}} \:+\:\mathrm{3}{y}^{\mathrm{3}} \:=\:\mathrm{200}\:\:\Rightarrow\:\:\left({x}^{\mathrm{2}} {y}^{\mathrm{2}} \right)_{\boldsymbol{{max}}} \:=\:? \\ $$
Commented by mr W last updated on 25/Jul/21
200=2x^3 +3y^3 ≥2(√((2x^3 )(3y^3 )))=2(√6)(x^2 y^2 )^(3/4) ⇒x^2 y^2 ≤(((200)/(2(√6))))^(4/3) =((500)/( ((45))^(1/3) ))=max
$$\left.\mathrm{200}=\mathrm{2}{x}^{\mathrm{3}} +\mathrm{3}{y}^{\mathrm{3}} \geqslant\mathrm{2}\sqrt{\left(\mathrm{2}{x}^{\mathrm{3}} \right)\left(\mathrm{3}{y}^{\mathrm{3}} \right.}\right)=\mathrm{2}\sqrt{\mathrm{6}}\left({x}^{\mathrm{2}} {y}^{\mathrm{2}} \right)^{\frac{\mathrm{3}}{\mathrm{4}}} \\ $$$$\Rightarrow{x}^{\mathrm{2}} {y}^{\mathrm{2}} \leqslant\left(\frac{\mathrm{200}}{\mathrm{2}\sqrt{\mathrm{6}}}\right)^{\frac{\mathrm{4}}{\mathrm{3}}} =\frac{\mathrm{500}}{\:\sqrt[{\mathrm{3}}]{\mathrm{45}}}={max} \\ $$
Commented by mathdanisur last updated on 25/Jul/21
Thank you Sir
$${Thank}\:{you}\:{Sir} \\ $$
Commented by mathdanisur last updated on 25/Jul/21
Ser, you used the condition that the variables are not negative, but the condition is not given
$${Ser},\:{you}\:{used}\:{the}\:{condition}\:{that}\:{the} \\ $$$${variables}\:{are}\:{not}\:{negative},\:{but}\:{the} \\ $$$${condition}\:{is}\:{not}\:{given} \\ $$
Commented by mr W last updated on 25/Jul/21
for local maximum, both x and y are positive. in fact this question is not a good one, since (x^2 y^2 ) has neither a maximum nor a minimum.
$${for}\:{local}\:{maximum},\:{both}\:{x}\:{and}\:{y}\:{are} \\ $$$${positive}. \\ $$$${in}\:{fact}\:{this}\:{question}\:{is}\:{not}\:{a}\:{good}\:{one}, \\ $$$${since}\:\left({x}^{\mathrm{2}} {y}^{\mathrm{2}} \right)\:{has}\:{neither}\:{a}\:{maximum} \\ $$$${nor}\:{a}\:{minimum}. \\ $$
Answered by Olaf_Thorendsen last updated on 25/Jul/21
2x^3 +3y^3 = 200 x = 200^(1/3) ((cos^(2/3) t)/2^(1/3) ), y = 200^(1/3) ((sin^(2/3) t)/3^(1/3) ) x^2 y^2 = ((200^(4/3) )/(2^(2/3) 3^(2/3) ))cos^(4/3) t.sin^(4/3) t x^2 y^2 = ((200^(4/3) )/(2^(2/3) 3^(2/3) 2^(4/3) ))sin^(4/3) (2t) x^2 y^2 = ((2^2 .5^(8/3) )/3^(2/3) )sin^(4/3) (2t) (d/dt)(x^2 y^2 ) = ((2^2 .5^(8/3) )/3^(2/3) )(2.(4/3).cos(2t)sin^(1/3) (2t)) (d/dt)(x^2 y^2 ) = ((2^5 .5^(8/3) )/3^(5/3) )(cos(2t)sin^(1/3) (2t)) (d/dt)(x^2 y^2 ) = 0 ⇔ 2t = ±((kπ)/2)+2kπ or 2t = ±kπ • For t = (π/4) : x = 200^(1/3) ((cos^(2/3) (π/4))/2^(1/3) ) = 2^(1/3) 5^(2/3) y = 200^(1/3) ((sin^(2/3) (π/4))/3^(1/3) ) = ((2^(2/3) 5^(2/3) )/3^(1/3) ) x^2 y^2 = (2^(1/3) 5^(2/3) )^2 (((2^(2/3) 5^(2/3) )/3^(1/3) ))^2 = ((2^2 5^(8/3) )/3^(2/3) )
$$\mathrm{2}{x}^{\mathrm{3}} +\mathrm{3}{y}^{\mathrm{3}} \:=\:\mathrm{200} \\ $$$${x}\:=\:\mathrm{200}^{\mathrm{1}/\mathrm{3}} \frac{\mathrm{cos}^{\mathrm{2}/\mathrm{3}} {t}}{\mathrm{2}^{\mathrm{1}/\mathrm{3}} },\:{y}\:=\:\mathrm{200}^{\mathrm{1}/\mathrm{3}} \frac{\mathrm{sin}^{\mathrm{2}/\mathrm{3}} {t}}{\mathrm{3}^{\mathrm{1}/\mathrm{3}} } \\ $$$${x}^{\mathrm{2}} {y}^{\mathrm{2}} \:=\:\frac{\mathrm{200}^{\mathrm{4}/\mathrm{3}} }{\mathrm{2}^{\mathrm{2}/\mathrm{3}} \mathrm{3}^{\mathrm{2}/\mathrm{3}} }\mathrm{cos}^{\mathrm{4}/\mathrm{3}} {t}.\mathrm{sin}^{\mathrm{4}/\mathrm{3}} {t} \\ $$$${x}^{\mathrm{2}} {y}^{\mathrm{2}} \:=\:\frac{\mathrm{200}^{\mathrm{4}/\mathrm{3}} }{\mathrm{2}^{\mathrm{2}/\mathrm{3}} \mathrm{3}^{\mathrm{2}/\mathrm{3}} \mathrm{2}^{\mathrm{4}/\mathrm{3}} }\mathrm{sin}^{\mathrm{4}/\mathrm{3}} \left(\mathrm{2}{t}\right) \\ $$$${x}^{\mathrm{2}} {y}^{\mathrm{2}} \:=\:\frac{\mathrm{2}^{\mathrm{2}} .\mathrm{5}^{\mathrm{8}/\mathrm{3}} }{\mathrm{3}^{\mathrm{2}/\mathrm{3}} }\mathrm{sin}^{\mathrm{4}/\mathrm{3}} \left(\mathrm{2}{t}\right) \\ $$$$\frac{{d}}{{dt}}\left({x}^{\mathrm{2}} {y}^{\mathrm{2}} \right)\:=\:\frac{\mathrm{2}^{\mathrm{2}} .\mathrm{5}^{\mathrm{8}/\mathrm{3}} }{\mathrm{3}^{\mathrm{2}/\mathrm{3}} }\left(\mathrm{2}.\frac{\mathrm{4}}{\mathrm{3}}.\mathrm{cos}\left(\mathrm{2}{t}\right)\mathrm{sin}^{\mathrm{1}/\mathrm{3}} \left(\mathrm{2}{t}\right)\right) \\ $$$$\frac{{d}}{{dt}}\left({x}^{\mathrm{2}} {y}^{\mathrm{2}} \right)\:=\:\frac{\mathrm{2}^{\mathrm{5}} .\mathrm{5}^{\mathrm{8}/\mathrm{3}} }{\mathrm{3}^{\mathrm{5}/\mathrm{3}} }\left(\mathrm{cos}\left(\mathrm{2}{t}\right)\mathrm{sin}^{\mathrm{1}/\mathrm{3}} \left(\mathrm{2}{t}\right)\right) \\ $$$$\frac{{d}}{{dt}}\left({x}^{\mathrm{2}} {y}^{\mathrm{2}} \right)\:=\:\mathrm{0} \\ $$$$\Leftrightarrow\:\mathrm{2}{t}\:=\:\pm\frac{{k}\pi}{\mathrm{2}}+\mathrm{2}{k}\pi\:\mathrm{or}\:\mathrm{2}{t}\:=\:\pm{k}\pi \\ $$$$\bullet\:\mathrm{For}\:{t}\:=\:\frac{\pi}{\mathrm{4}}\:: \\ $$$${x}\:=\:\mathrm{200}^{\mathrm{1}/\mathrm{3}} \frac{\mathrm{cos}^{\mathrm{2}/\mathrm{3}} \frac{\pi}{\mathrm{4}}}{\mathrm{2}^{\mathrm{1}/\mathrm{3}} }\:=\:\mathrm{2}^{\mathrm{1}/\mathrm{3}} \mathrm{5}^{\mathrm{2}/\mathrm{3}} \\ $$$${y}\:=\:\mathrm{200}^{\mathrm{1}/\mathrm{3}} \frac{\mathrm{sin}^{\mathrm{2}/\mathrm{3}} \frac{\pi}{\mathrm{4}}}{\mathrm{3}^{\mathrm{1}/\mathrm{3}} }\:=\:\frac{\mathrm{2}^{\mathrm{2}/\mathrm{3}} \mathrm{5}^{\mathrm{2}/\mathrm{3}} }{\mathrm{3}^{\mathrm{1}/\mathrm{3}} } \\ $$$${x}^{\mathrm{2}} {y}^{\mathrm{2}} \:=\:\left(\mathrm{2}^{\mathrm{1}/\mathrm{3}} \mathrm{5}^{\mathrm{2}/\mathrm{3}} \right)^{\mathrm{2}} \left(\frac{\mathrm{2}^{\mathrm{2}/\mathrm{3}} \mathrm{5}^{\mathrm{2}/\mathrm{3}} }{\mathrm{3}^{\mathrm{1}/\mathrm{3}} }\right)^{\mathrm{2}} =\:\frac{\mathrm{2}^{\mathrm{2}} \mathrm{5}^{\mathrm{8}/\mathrm{3}} }{\mathrm{3}^{\mathrm{2}/\mathrm{3}} } \\ $$
Commented by mr W last updated on 25/Jul/21
please check before saying! but 50(25^(1/3) ) is wrong! and ((2^2 5^(8/3) )/3^(2/3) ) is right!
$${please}\:{check}\:{before}\:{saying}! \\ $$$${but}\:\mathrm{50}\left(\mathrm{25}^{\frac{\mathrm{1}}{\mathrm{3}}} \right)\:{is}\:{wrong}! \\ $$$${and}\:\frac{\mathrm{2}^{\mathrm{2}} \mathrm{5}^{\mathrm{8}/\mathrm{3}} }{\mathrm{3}^{\mathrm{2}/\mathrm{3}} }\:{is}\:{right}! \\ $$
Commented by mathdanisur last updated on 25/Jul/21
Thank you Ser But answer: 50(25^(1/3) )
$${Thank}\:{you}\:{Ser} \\ $$$${But}\:{answer}:\:\mathrm{50}\left(\mathrm{25}^{\frac{\mathrm{1}}{\mathrm{3}}} \right) \\ $$
Commented by mathdanisur last updated on 25/Jul/21
Thankyou Ser
$${Thankyou}\:{Ser} \\ $$

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