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

Note by Abhishek Alva
1 year ago

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Is it 3456/3125. ?

Naitik Sanghavi - 1 year ago

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please write the solution

Abhishek Alva - 1 year ago

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Am i correct?

Naitik Sanghavi - 1 year ago

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@Naitik Sanghavi yes u are write the solution

Abhishek Alva - 1 year ago

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@Abhishek Alva Apply Am gm on x,x,2y/3,2y/3,2y/3.

Naitik Sanghavi - 1 year ago

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@Naitik Sanghavi 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 numbers

Abhishek Alva - 1 year ago

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@Abhishek Alva i not getting this problem

Abhishek Alva - 1 year ago

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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 year ago

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i guess u used the cauchy schwarz inequality

Abhishek Alva - 1 year ago

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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 mistake @Ayush Rai.

Naitik Sanghavi - 1 year ago

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

Ayush Rai - 1 year ago

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@Ayush Rai I already did iny other comment

Naitik Sanghavi - 1 year ago

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