Question 965124
Question 1 is a rectangular shape that can be analyzed.  Question 2 shape is unspecified so is incomplete.


x and y dimensions of the rectangle.
p for perimeter:
{{{p=2x+2y}}} and {{{xy=48}}}
-
{{{p=2x+2(48/x)}}}
{{{p=2x+96/x}}}


The only method for minimizing or maximizing the perimeter seems to be derivatives.


{{{dp/dx=2+96*(d/dx)(x^(-1))}}}
{{{2+96(-1)(x^-2)}}}
{{{2-96/x^2}}}


Finding extreme values, set derivative to zero.  Continue other steps.
{{{dp/dx=2-96/x^2=0}}}
{{{(2x^2-96)/x^2=0}}}
Denominator cannot be zero, and for derivative be zero, the NUMERATOR must be equated to zero.


{{{2x^2-96=0}}}
{{{x^2-48=0}}}
{{{x^2=48}}}
{{{x=sqrt(48)}}}, the positive value only.

-
{{{x=sqrt(2*24)=sqrt(2*2*6*2)=sqrt(2*2*2*2*3)}}}
{{{highlight(x=4*sqrt(3))}}}
-
and from the original area equation, obviously {{{y=4*sqrt(3)}}}, making the area shape as a SQUARE shape.  The values for MINIMUM perimeter.


Note that you will find this is for MAXIMUM area.  Check about the area equation and you should find no minimum.




Here is a graph for the PERIMETER equation, in one or either variable, here being {{{p=2x+48/x}}}, which shows a MINIMUM for the perimeter:
{{{graph(350,350,-8,39,-8,39,2x+96/x)}}}