What is the maximum value of \(x^2 y^3\) for positive numbers \(x\) and \(y\) satisfying \(x+y=2\)?

What is the maximum value of \(x^2 y^3\) for positive numbers \(x\) and \(y\) satisfying \(x+y=2\)?

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TopNewestIs it 3456/3125. ? – Naitik Sanghavi · 1 week, 2 days ago

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– Abhishek Alva · 1 week, 2 days ago

please write the solutionLog in to reply

– Naitik Sanghavi · 1 week, 2 days ago

Am i correct?Log in to reply

– Abhishek Alva · 1 week, 2 days ago

yes u are write the solutionLog in to reply

– Naitik Sanghavi · 1 week, 2 days ago

Apply Am gm on x,x,2y/3,2y/3,2y/3.Log in to reply

– Abhishek Alva · 1 week, 2 days ago

a/(a^3+b^2+c )+b/(b^3+c^2+a)+c/(c^3+a^2+b) if a+b+c=3 determine the largest value of the equation given that a b c are positive real numbersLog in to reply

– Abhishek Alva · 1 week, 2 days ago

i not getting this problemLog in to reply

I got the answer to be \(\dfrac{676\sqrt{26}}{3125}\) which is very close to your answer.First we get the max. value of \(xy\) to be \(1\) and \(x^2+y^2\) to be \(2\) by \(AM-GM\) inequality.Now we must find the max value of \(2x+3y.\)We have an inequality which states that \(ax+by\leq \sqrt{a^2+b^2}\times\sqrt{x^2+y^2}.\)Well I dont know the proof for this inequality.Hence, \(2x+3y\leq\sqrt{2^2+3^2}\times\sqrt 2=\sqrt{26}.\) Next we apply AM-GM inequality for \(2x+3y.\)We split the terms and apply the inequality\(.\dfrac{x+x+y+y+y}{5}\geq\sqrt[5]{x^2y^3}\Rightarrow \dfrac{2x+3y}{5}\geq\sqrt[5]{x^2y^3}.\)Therefore,\(\dfrac{\sqrt{26}}{5}\geq\sqrt[5]{x^2y^3}.\)Now raise both sides by power of \(5.\)We get the maximum value of \(x^2y^3\) to be \(\boxed{\dfrac{676\sqrt{26}}{3125}}\)\(.\) – Ayush Rai · 1 week ago

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– Abhishek Alva · 1 week ago

i guess u used the cauchy schwarz inequalityLog in to reply

@Ayush Rai. – Naitik Sanghavi · 1 week ago

When you applied am gm on x,x,y,y,y the equality holds when x=y which implies that max value of x²y³ is 1.I think this was ur mistakeLog in to reply

– Ayush Rai · 1 week ago

But it is very close to the answer.Can u point the error?Log in to reply

– Naitik Sanghavi · 1 week ago

I already did iny other commentLog in to reply