A piece of wire 11 m long is cut into two pieces. One piece is bent into a square and the other is bent into an equilateral triangle. How much wire should be used for the square in order to maximize and minimize the total area?

2 Answers
May 31, 2018

For maximum area, all of the wire should be used to construct the square.
The minimum total area is obtained when # (44)/(4+3sqrt 3)\ "m"~~4.78\ "m"# is used for the square.

Explanation:

Let's say that #x# m is used for the square.

Then the length of the side of the square #= x/4# m and its area #=x^2/16\ "m"^2#

On the other hand, the side of the equilateral triangle will then be #a= (11-x)/3\ "m"# and its area will be #sqrt 3/4a^2 = sqrt3/36(11-x)^2\ "m"^2#

Hence the total area is #A\ "m"^2# where

#A = x^2/16+ sqrt3/36(11-x)^2#

To extremize this, we find the derivatives

#(dA)/dx = x/8-sqrt 3/18 (11-x)#

and

#(d^2A)/dx^2 = 1/8+sqrt3/18>0#

Thus, the extremum is at

#(dA)/dx = 0 implies#
# x/(11-x) = (4sqrt 3)/9 = 4/(3sqrt 3)implies#
#x/11 = (4)/(4+3sqrt 3)implies#
#x = (44)/(4+3sqrt 3)~~4.78#

Since #(d^2A)/dx^2 >0# this is a minimum.

Thus the minimum area of #~~3.29\ "m"^2# will be obtained when
#4.78\ "m"# is used for the square.

This is the only extremum for #x in (0,11)# and so the maximum occurs at a boundary of the allowed values. It is easy to see that the absolute maximum is for #x= 11#.

This is understandable, because for a given perimeter, a square has more area than a triangle - hence to get the largest area we need to use all of the wire to construct the square.

May 31, 2018

Minimum Area occurs is we use #12sqrt(3)-16 ~~ 4.786609 \ m# for the square.

Maximum Area occurs is we use #11 \ m# for the square.

Explanation:

Let us set up the following variables:

# { (t, "Length of a side of the triangle",), (l, "Length of a side of the square",), (P_t, "Perimeter of a side of the triangle",=4t), (P_s, "Perimeter of a side of the square",=4s), (A_t, "Area of the triangle",), (A_s, "Area of the square",), (A, "Total Area",=A_t+A_s) :} #

Our aim is to find #A(t,l)#, as a function of a single variable and to maximize the total area, #A#, wrt that variable (It won't matter which variable we do this with as we will get the same result). ie we want a critical point of #A# wrt the variable.

The total perimeter is that of #4# sides of a square and the #3# sides of the triangle; we are told that this perimeter is #11# (m)

# P_s + P_t =11 => 4l+3t = 11 #
# :. t = 11 #

Similarly, the total Area is that of a square and the triangle:

# A = A_s+A_t#

# \ \ \ = (l)(l) + 1/2(t)(t)sin60^o #

# \ \ \ = l^2 + 1/2t^2 \ sqrt(3)/2 #

# \ \ \ = l^2 + sqrt(3)/4 t^2 #

# \ \ \ = l^2 + sqrt(3)/4 ((11-4l)/3)^2 #

# \ \ \ = l^2 + sqrt(3)/4 (11-4l)^2/9 #

# \ \ \ = l^2 + sqrt(3)/36 (11-4l)^2 #

We now have the Area, #A#, as a function of a single variable #l#, so differentiating wrt #t# we get:

# (dA)/(dl) = 2l + 2sqrt(3)/36 (11-4l)(-4) #

# \ \ \ \ \ \ = 2l -(2sqrt(3))/9 (11-4l) #

At a critical point we have #(dA)/(dl) =0 => 2l -(2sqrt(3))/9 (11-4l) = 0#

# :. l - sqrt(3)/9 (11-4l) = 0#

# :. 9l - sqrt(3) (11-4l) = 0#

# :. 9l - 11sqrt(3) +4sqrt(3)l = 0#

# :. (9 +4sqrt(3))l = 11sqrt(3)#

# :. l = (11sqrt(3))/(9 +4sqrt(3)) #

# :. l = (11sqrt(3))/(9 +4sqrt(3))* (9 -4sqrt(3))/(9 -4sqrt(3))#

# :. l = (99sqrt(3)-44*3)/(81-16*3)#

# :. l = (99sqrt(3)-132)/(81-48)#

# :. l = (99sqrt(3)-132)/(33)#

# :. l = 3sqrt(3)-4 ~~ 1.196152#

With this value of #l# we have, for the square, and triangle:

# {: (P_s, = 4l, = 12sqrt(3)-16, ~~ 4.786609 \ m), (P_t, = 11-4l, = 27-12sqrt(3), ~~ 6.215390 \ m) :} #

Similarly, if sought, we could compute:

# {: (A_s, = l^2, = 43-24sqrt(3), ~~ 1.430780 \ m^2), (A_t, = sqrt(3)/36 (11-4l)^2, = (129sqrt(3))/4-54, ~~ 1.858638 \ m^2) :} #

We can visually verify that this corresponds to a minimum by looking at the graph of #y=A(l)# (we could also perform a second derivative test)

graph{x^2 + sqrt(3)/36*(11-4x)^2 [-10, 10, -5, 20]}

It should also be intuitive (and from the graph) that there is no maximum bound on the area, thus to maximize the area we should use all of the wire