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Can anyone help me with finding the maximum value of $$y = {(ab)(a+b)}$$ With that condition that $$ 100 = a^2 + b^2$$ and both $a,b$ are positive numbers. Any help appreciated. Thanks.

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    $\begingroup$ Since this is a site that encourages and helps with learning, it is best if you show your own ideas and efforts in solving the question. Can you edit your question to add your thoughts and ideas about it? $\endgroup$ – 5xum Apr 1 '16 at 9:17
  • $\begingroup$ Use "Lagrange multipliers" $\endgroup$ – SomeOne Apr 1 '16 at 9:17
  • $\begingroup$ The answer might be an irrational one. $\endgroup$ – TheRandomGuy Apr 1 '16 at 10:00
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Since $a,b>0$ and hence $y>0$ you can equivalently maximize $y^2$ which you can write by the given constraint as $$y^2=(ab)^2(a+b)^2=(ab)^2(100+2ab)$$ so that $y^2$ (and hence $y$) is maximized whenever $ab$ is maximized. By the AM-GM inequality $$ab=\sqrt{(ab)^2}\overset{AM-GM}\le\frac{a^2+b^2}{2}=\frac{100}{2}=50$$ with equality if $a=b$, see here. Hence $a=b=\sqrt{50}$ with $$y_{\max}=\sqrt{50}^2(\sqrt{50}+\sqrt{50})=100\sqrt{50}$$


To use the method of Lagrange multipliers you can write this problem as \begin{align}\max_{a,b} {f(a,b)}&=(ab)(a+b)\\\mbox{s.t. } g(a,b)&=a^2+b^2-100=0\end{align} The Lagrange function is $$\mathcal L(a,b,λ)=f(a,b)-λg(a,b)$$ with partial derivatives \begin{align}\frac{\partial \mathcal L}{\partial a}&=2ab+b^2-2λa\\\frac{\partial \mathcal L}{\partial b}&=2ab+a^2-2λb\\\frac{\partial \mathcal L}{\partial λ}&=-α^2-b^2+100\end{align} If $(a_0,b_0)$ is a solution to the problem then there exists $λ_0$ such that $(a_0,b_0,λ_0)$ is a stationary point of $\mathcal L$. Now this is the point where you have sometimes (in this method) to "guess" a solution. Here an "obvious guess" is that $a=b$ (however, making this approach no better than others - see other answer or the first part of this answer) which gives \begin{align}3a^2-2λa&=0\\ 2a^2&=100 \end{align} which gives the (valid) solution $$(a_0,b_0,λ_0)=\left(\sqrt{50},\sqrt{50},\frac32\sqrt{50}\right)$$ as the first method.

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  • $\begingroup$ You're wrong. $y_{max} = 100\sqrt{50}$. Why've you taken $\sqrt{50}^2$? Shouldn't it be $\sqrt{50}$ in the parentheses where you define $y_{max}$. $\endgroup$ – TheRandomGuy Apr 1 '16 at 10:07
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Note that $(a-b)^2 \ge 0$ so that $ab \le \frac{a^2 +b^2}{2}$ with equality iff $a=b$

Also, $(a+b)^2 = a^2 +b^2 +2ab$. This should allow you to find least upper bound for both terms in the product.

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  • $\begingroup$ very neat solution! $\endgroup$ – Surb Apr 1 '16 at 9:22
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Other option, use the following change of variables (which always satisfies $a^2+b^2=100$): \begin{cases} a=10\cos t\\ b=10\sin t \end{cases} In terms of $t$, the objective functions equals: $$ y(t)=100\cos t \sin t(10\cos t+10\sin t) $$ Analyzing the derivative with respect to $t$ yields $$ y_{MAX}=500\sqrt{2} $$ with $t=\frac{\pi}{4}$, that is $$ \begin{cases} a=5\sqrt{2}\\ b=5\sqrt{2} \end{cases} $$

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