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I have the task: find maximum an minimum of $$f(x) = x_1(\pi - x_1)\sin x_2 + x_2 \cos x_1$$ on X where $$X = \{x\in R^2\ |\ x_1\in [0, \pi], x_2 \ge 0\}.$$ First thing i did was system : \begin{equation*} \begin{cases} \frac{\partial f(x)}{\partial x_1} = 0\\ \frac{\partial f(x)}{\partial x_2} = 0\\ \end{cases} \end{equation*} and i got \begin{equation*} \begin{cases} \sin x_2(\pi - 2x_1) - x_2\sin x_1 = 0\\ (\pi x_1 - x_1^2)\cos x_2 + \cos x_1 = 0\\ \end{cases} \end{equation*} and i didn't solve it .

Wolfram says: enter image description here

Simankov
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2 Answers2

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Hint

From the second equation, you can eliminate $\cos(x_2)$ and therefore $x_2$ as a function of $x_1$.

Plug it in the first equation and you have a single equation to solve for $x_1$.

More than likely, you will find several roots. I suppose that numerical methods would be required.

Claude Leibovici
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  • Then i get equation with arccos, roots... I can't solve it on paper. There is a classroom task. I think, there is another solution based on Lagrange methods or some. Thanks – Simankov May 03 '15 at 14:06
  • I agree that it is impossible to solve (at least to me). However, you can plot the first equation and locate the roots. Sorry for not being able to do more for you. I have the strange feeling that the minimum could correspond to $x_1=\pi$, $x_2=2\pi$ for which $f=-2\pi$. – Claude Leibovici May 03 '15 at 14:23
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I found that when $ x1=0$ and $x2>=0$ -> $f(x)=x2$, so we didn't have global maximum and that when $x1=π$ and $x2>=0$ -> $f(x)=−x2$ so we didn't have global minimum. Therefore answer is there aren't maximum and minimum

Simankov
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