# maximum area of rectangle inscribed in a circle using geometric programming

need to find maximum area of rectangle that can be inscribed in a circle of radius r but need to use geometric programming of optimization to this

for the maximum area the function is $xy$ (if x is length and y is breadth and radius is r ) and the constrain what i got is $(4r^2\cdot y^{-2})-(x^2\cdot y^{-2})=1$ since $x^2+y^2=(2r)^2$

but the degree of difficult for above is "$-1$" So Are the above equations are correct or am i missing a logic if they are wrong what should be the equations if they are correct how should i proceed further

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Your problem is that you try to model "inscribed rectangle" by $x^2+y^2=4r^2$. You should model it by $x^2+y^2 \leq 4r^2$ instead, because you will have equality for the maximal rectangle anyway.

For the formulation in the standard form of a geometric program, I would use $x_1:=\frac{x}{2r}$ and $x_2:=\frac{y}{2r}$. The area of the rectangle is then given by $A=4r^2 x_1x_2$, and we must solve the following geometric program:

Minimize $x_1^{-1}x_2^{-1}$ subject to $x_1^2+x_2^2 \leq 1$.

Because of the uniqueness properties of the solution and the symmetry of the problem statement with respect to interchanging $x_1$ and $x_2$, the solution must satisfy $x_1=x_2$. It is also obvious that the solution must satisfy $x_1^2+x_2^2 = 1$. Hence we get $x_1=x_2=\sqrt{1/2}$ and $A=2r^2$.

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A=4r^2xy how? i was not able to get it and why we need take x=y=sqrt(1/2) – karthik patruni Feb 13 '12 at 11:32
I rewrote the answer to clarify the unclear points. Considering the limited feedback to your question, you might try asking similar questions on scicomp.stackexchange.com or or-exchange.com – Thomas Klimpel Feb 13 '12 at 18:12

You are right that $A=xy$ with the constraint $x^2+y^2=4r^2$. You need to solve the constraint equation to remove one variable. So $x=\sqrt {4r^2-y^2}$. Now plug this into your area equation to get $A=y\sqrt {4r^2-y^2}$. Then $\frac {dA}{dy}=\sqrt{4r^2-y^2}-\frac {y^2}{\sqrt{4r^2-y^2}}$. Setting this to zero, we get at the maxiumum $2y^2=4r^2$ or $y=r\sqrt 2$. Then we find that $x=r\sqrt 2$ as well.

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