Question

Could use the extra credit :)
Prove that for a rectangle of fixed perimeter, the square gives the largest area and that given a fixed area, the square gives the smallest perimeter.

I proved the first, don't know how to prove the second. It's a corollary.
P = 2l + 2w
P = 2l+ 2x (given that w=x)
P-2x =2l
l = (P-2x)/2

A= lw
=((P-2x)/2)(x)
=(1/2)(Px-2x^2)
dA/dt = (1/2)(P-4x)
0 =(1/2)(P-4x) --- Stationary points = max or min
0=P-4x
x=P/4
For the area of a fixed P in a rectangle to be largest, each side must be 1/4 the perimeter, therefore a square.

Help me prove the 2nd pt of the proof plz

Answer 1

in your proof above, you should also show that the 2nd derivative is negative for the value of x that makes the 1st derivative zero. that will prove it is a maximum and not a minimum.
the 2nd proof is very similar:
A=lw
so l=A/w
P=2l+2w=2A/w + w
so Pw = 2A + w^2
find dP/dw and then see what value of w makes this zero, etc - same principal as you followed above.