Let the length of the large rectangle enclosing the other three rectangles have dimensions #x# and #y#
Therefore #x y#=#4000#.......#[1]# The total length of the fencing must be the length enclosing the large rectangle i.e, #2x#+#2y# + the length of two cross pieces whose length is #x# [ choosing to divide the large rectangle along it's #y# length]
Therefore total length of fencing is #4x+2y#.........#[2]#. From .....#[1]# , #y=4000/x# and substituting this value of #x# into.......#[2]# will give us , length of fencing, say #L# =#4x+[2][4000/x]# Simplifying, #L=x+2000/x#, we need to differentiate this expression and set the derivitive = #0# to find max/min.
#[dL]/dx=1 - 2000/x^2# =#0# , which gives #x^2=2000#, and simpilfying, #x=10sqrt20# [ ignoring the the negative square root for #x#, since we cannot have a negative length] ]
substituting this value of #x# into......#[1]# will give a #y# value of #20sqrt20#
Therefore, from......#[2]# total length of fencing #L# =#4x+2y#=#4[10sqrt20]+ 2[20sqrt20# = # 80sqrt20#'
To find if the value of #x# minimises the the length of fencing to enclose the area of the rectangle we must check the second derivative, which if positive, will indicate a local minimum of the function #L#
#dL/dx=1-2000/x^2# , so #d/dx[dL/dx]#= #2000/x^3# which is positive when #x# =#10sqrt20# and so this value of #x# will minimise #L# . Hope this was helpful.
