Question #ef706

2 Answers
Apr 8, 2015

The optimal Area to Perimeter ratio is achieved by a circle.

Removal of any part of the circle's perimeter (circumference) to construct any other shape is counter productive.

Use the entire 36 meters for the circle (make your square #0xx0# meters).

Apr 8, 2015

This minimization problem turns into a minimization on a closed interval. I haven't found a general post on the closed interval method. (Maybe the link owl will spot one.)

Let #x# = the piece formed into a square.
Then the sides of the square are each of length #x/3# and the area of the square is #x^2/16#

The piece formed into a circle measures #36-x#. That will be the circumference of the circle. We need the area as a function of this circumference.

For a circle, #A= pi r^2# and #C=2 pi r# We have #C=36-x#, so that entails that #r=(36-x)/(2 pi)# and #A = pi ((36-x)/(2 pi))^2#

The total area will be

#T= x^2/16 + (36-x)^2/(4 pi) # for #0 <=x <= 36#

Minimize the function of the interval:

#T' = x/8 - (36-x)/(2pi) = (pi x-144 + 4x)/(8x)#

The zero of #T'# occurs where #pi x-144 + 4x = 0#

The critical numbers are #x=144/(4+pi)# and #x=0#.

Calculate the Total area for the 3 values of #x#,

#x=0#, #x=144/(4+ pi)# and #x= 36# to see where the minimum occurs.