Let us call X the length of the rectangle and x the width if it.
Perimeter is 2X+2x so we have 2X+2x=116.
Which means X=(116-2x)/2 which means X=58-x
Surface (S) is X*x
So we can deduce:
S=X*x
S=(58-x)*x
S=58x-[tex] x^{2} [/tex]
We are looking to maximise.
This happens when the derivate equals 0
Derivate is 58-2x
58-2x=0 if x=29
So the surface is maximized when x=29
Which means X=29 as well (X=58-x)
Got it?
