Question 1099767
I see the first question as a quadratic function question,
although you need to know enough geometry to understand what a rectangle is.
The second question, looks more like a calculus question
 
a) {{{drawing(600,250,-2,10,-1,4,
rectangle(0,0,8,3),red(line(-2,0,11,0)),
locate(2.8,0,red(barn)),locate(4.2,0,red(wall)),
locate(0.05,1.8,x),locate(7.8,1.8,x),locate(3,3,2100m-2x)
)}}} The area {{{y}}} of that pen (in square meters) is {{{y=x(2100-2x)}}} .
You should recognize that as a quadratic function,
which graphs as a parabola, looking like this:
{{{graph(200,200,-0.1,0.9,-0.1,0.9,4x-5x^2)}}} , with two zeros and a maximum exactly midway between them.
Re-writing it as {{{y=2x(1050-x)}}} shows you clearly that
{{{y=0}}} for {{{x=0}}} and {{{x=1050}}} ,
{{{y>0}}} in between, for {{{0<x<1050}}} ,
and the maximum is at {{{x=1050/2=525}}}
The dimensions that maximize the area are
{{{x=highlight(525m)}}} for the length of each of the two fencing sides attached to the barn wall, and
{{{2100m-2x=2100m-2(525m)=2100m-1050m=highlight(1050m)}}} for the length of of the fencing side parallel to the barn wall.
 
b) If we design something with four identical pens, like this:
{{{drawing(600,250,-2,10,-1,4,
rectangle(0,0,8,3),rectangle(2,0,8,3),
rectangle(4,0,6,3),
red(line(-2,0,11,0)),
locate(2.8,0,red(barn)),locate(4.2,0,red(wall)),
locate(0.05,1.8,x),locate(0.9,3,y)
)}}} , where {{{x}}} and {{{y}}} are lengths in m,
we know that {{{x*y=400m^2}}} <---> {{{y=400/x}}} ,
and that makes the total length of fence needed, {{{y}}} , in m
{{{y=5x+4(400/x)=5x+1600/x}}} .
Maybe you are supposed to use a graphing calculator to find that the minimum for {{{y}}} happens at approximately {{{x=17.888544}}} ,
and you could tell the farmer to use 5 {{{17.9m}}} length of fencing perpendicular to the barn wall,
attached to a {{{400m^2/"17.9 m"=approximately}}}{{{22.35m}}} length parallel to the wall.
 
Using calculus, you would find that the derivative is
{{{dy/dx=5-1600/x^2=5-(40/x)^2}}} .
As {{{dy/dx<0}}} for {{{-40/sqrt(5)<x<40/sqrt(5)}}} ,
{{{dy/dx=0}}} for {{{x=40/sqrt(5)}}} , and {{{dy/dx>0}}} for {{{x>40/sqrt(5)}}} ,
the function decreases for {{{x>0}}} to a minimum at {{{x=40/sqrt(5)=approximately}}}{{{17.888544}}} , and increases for {{{x>40/sqrt(5)}}} .
 
If there is another alternative approach, let me know.