Maximum value of $(ab)(a+b)$ given $a^2 + b^2 = 100$

Question

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.

Answer 1

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 $$y_{\max}=\sqrt{50}^2(\sqrt{50}+\sqrt{50})=100\sqrt{50}$$

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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.

Answer 2

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.

Answer 3

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 $$ \begin{cases} a=5\sqrt{2}\\ b=5\sqrt{2} \end{cases} $$