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Mathematics 11 Online
OpenStudy (anonymous):

Of all rectangles with perimeter 10 m, which one has the maximum area?

OpenStudy (anonymous):

the one that is a square

OpenStudy (anonymous):

but how do you get to that?

OpenStudy (cwrw238):

do you know some calculus?

OpenStudy (anonymous):

by symmetry

OpenStudy (anonymous):

of course @cwrw238

OpenStudy (anonymous):

you can call one side \(x\) and the other side \(y\) but you cannot tell \(x\) from \(y\) and so there is no difference between them

OpenStudy (anonymous):

okay

OpenStudy (anonymous):

or you can call one side \(x\) and the other side \(y\) then you kbnow \[2x+2y=10\] or \[y=5-x\] and maximize \[x(5-x)=5x-x^2\] max is at the vertex, namely \(-\frac{b}{2a}=2.5\)

OpenStudy (anonymous):

but really it is more obvious than that you have a rectangle largest area is a square by symmetry

OpenStudy (anonymous):

oh I see. Thanks for that. I guess that's what i really needed was to just to see it for myself.

OpenStudy (anonymous):

yw

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