OpenStudy (anonymous):

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

4 years ago
OpenStudy (anonymous):

the one that is a square

4 years ago
OpenStudy (anonymous):

but how do you get to that?

4 years ago
OpenStudy (cwrw238):

do you know some calculus?

4 years ago
OpenStudy (anonymous):

by symmetry

4 years ago
OpenStudy (anonymous):

of course @cwrw238

4 years ago
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

4 years ago
OpenStudy (anonymous):

okay

4 years ago
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 \(-\frac{b}{2a}=2.5\)

4 years ago
OpenStudy (anonymous):

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

4 years ago
OpenStudy (anonymous):

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

4 years ago
OpenStudy (anonymous):

yw

4 years ago