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We have a wire mesh of 1000 m to fence 2 regions, one circular and one square. Say how should the mesh should be cut to: a) The sum of the areas of both fenced regions is maximum. b) The sum of the areas of both fenced regions is minimum.

I don't know if I should use Lagrange Multpliers or if it's only a one variable calculus problem. Any hint or idea would be very appreciated. Thank you.

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Yes, you can use Lagrange multipliers and yes, it can be expressed as a $1$-variable problem. Your pick.

Let $x$ be the radius of the circle and $y$ the side of the square. We have the constraint $$2\pi x+4y=1000\tag{1}.$$ We want to maximize/minimize $$\pi x^2+y^2\tag{2}$$ subject to Condition (1).

Now use Lagrange multipliers. Things should go smoothly. One must not forget to check the endpoints $x=0$ and $y=0$.

Or else we can use (1) to say solve for $y$ in terms of $x$, and substitute for $y$ in (2). We then have a one-variable problem, to be solved in the usual introduction to calculus way, or some other way. We get a quadratic in $x$, with somewhat messy coefficients.

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  • $\begingroup$ Using the second method I got $f(x)=(\pi+\frac{\pi^2}{4})x^2-250\pi x + 250^2$, so $f'(x)=(2\pi+\frac{\pi^2}{2})x-250\pi$. If $f'(x)=0$, then $x=\frac{250\pi}{2\pi+\frac{\pi^2}{2}}=\frac{500}{4+\pi}$. Since $f''(x)=2\pi+\frac{\pi^2}{2}>0$, $\frac{500}{4+\pi}$ is a minimum. But, what about the maximum? Since $f$ is quadratic, it can have only one critical point, not two. How can I find the maximum? $\endgroup$ – Twink Sep 21 '13 at 19:44
  • $\begingroup$ Remember I repeated a couple of times that one had to check for endpoint maxima/minima. In this case, we get maximum area by giving everything to the circle. $\endgroup$ – André Nicolas Sep 21 '13 at 22:14

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