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Question is to find Finding Triangle with constant perimeter and largest area by method of lagrange multiplier .

What i have done is that i have firstly taken $x+y+z=2k$ , where x,y,z are sides of triangle..k is any constant

Then i use Heron's formula as $\sqrt{s(s-x)(s-y)(s-z)}$ , where $s = (x +y+z)/2$ ....

Since Area = $\sqrt{s(s-x)(s-y)(s-z)}$

So substituting values of $s$ and replacing $z$ by $2k-x-y$ (to make it to two variable problem ) i finally get

$f(x,y) = k(k-x)(k-y)(x+y-k)$ ...($A$ is squared so as to be easy easy derivatives)

And my constraint equation is $g=x+y+z-2k=0 $ .....But problem here is that constraint consists z also . So i feel stuck to use LAGRANGE MULTIPLIER Method.....Can any1 help me furthure what to do from here .THANKS

godonichia
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1 Answers1

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There is no reason to eliminate $z$. The constraint is $x+y+z=2k$, and the objective function is $(k-x)(k-y)(k-z)$.

André Nicolas
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  • i would be 3 variable problem then .. – godonichia Nov 15 '14 at 01:52
  • Three variables are fine. There is a net symmetry gain that more than makes up for the extra variable. – André Nicolas Nov 15 '14 at 01:53
  • ok i vl try ur way .but if i want to proceed from where i have left off then what should i do – godonichia Nov 15 '14 at 01:59
  • as per ur prescription i have got $f_x$ = $(k-y)(k-z) + \lambda $ , $f_y$ = $(k-x)(k-z) + \lambda$ , $f_z$ = $(k-x)(k-y)$ and the constraint equation (But i doesnt seem to need it here ) . Now equating $ \lambda $ from three equations , one by one ,and cancelling common factors as say from 1st two equations $(k-y)(k-z)$ = $(k-x)(k-z)$ ..so$(k-y)=(k-x)$ and so $x=y$ and similarly i get $x=y=z$ . AM I CORRECT ?? – godonichia Nov 15 '14 at 02:23
  • Things are basically right. You have sign errors for $f_x$, $f_y$, $f_z$ but they make no difference. The cancellations are not quite right either, when you cancel $k-z$ you are throwing away a solution. – André Nicolas Nov 15 '14 at 02:33
  • that sign i have deliberately not written here bcos it makes no difference , but can u explain other point – godonichia Nov 15 '14 at 02:36
  • From $(k-y)(k-z)=(k-x)(k-z)$ you conclude that $k-y=k-x$ and therefore $y=x$. But maybe $k-z=0$, that would also make the equality true. Fairly quickly however, from the fact that the product is also equal to $(k-x)(k-y)$, we conclude that then one of $x$ or $y$ is equal to $k$, giving a degenerate triangle. Thus the cancellation turned out to not lead to the wrong conclusion. However, as a matter of general carefulness, we cannot freely cancel. – André Nicolas Nov 15 '14 at 02:44
  • If $(k-x)(k-y)=0$ then one of $k-x$ or $k-y$ is $0$. Suppose $k-x=0$, so $x=k$. Recall we had $z=k$. But the three sides add up to $2k$. So the third side $y$ is $0$. A "triangle" with a side equal to $0$ is called degenerate. – André Nicolas Nov 15 '14 at 03:11
  • Yes Thanks a lot ! – godonichia Nov 15 '14 at 03:13
  • You are welcome. As I said earlier, your calculation was basically right. However, in another situation, the root you "threw away" might have been the relevant one. – André Nicolas Nov 15 '14 at 03:16
  • Yes i will take care of that in future .i basically cancelled it thinking that the difference can never be zero of that term as they rpresent sides and perimeter . – godonichia Nov 15 '14 at 03:38
  • There will always be justification for throwing root in every situation like this ? – godonichia Nov 15 '14 at 03:39
  • If u have time pls see this http://math.stackexchange.com/questions/1002316/multiple-integration-questions-and-my-attempt – godonichia Nov 15 '14 at 03:42