Bob wants to cut a wire that is 60 cm long into two pieces. Then he wants to make each piece into a square. Determine how the wire should be cut so that the total area of the two squares is a small as possible?

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Answer 1 · 2016-12-02T14:10:35

The 60 cm long wire should be cut so that you have 2 lengths of 30 cm each.

Let 1 of the cut pieces total length be #x#
Then the other piece is length #60-x#

Let the area for square 1 be #A_1#
Let the area for square 2 be #A_2#
Let the sum of the areas be #A_s#

All sides of a square are of equal length so:

#A_1=(x/4)^2#

#A_2=((60-x)/4)^2#

#A_s=A_1+A_2=(x/4)^2+((60-x)/4)^2#

#A_s= x^2/16+(x^2-120x+3600)/16#

#A_s= (2x^2)/16- 120/16x+3600/16#

#A_s=1/8x^2-15/2x+225#
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This is a quadratic and as the #x^2# term is positive it is of general shape of #uu#

Thus the minimum area (#A_s#) is at the vertex. Let me show you a trick to determining the vertex #x# value.

Write as #A_s=1/8(x^2-(8xx15)/2x)+225#

#A_s=1/8(x^2-color(red)((cancel(8)^4xx15)/(cancel(2)^1))x)+225#

#color(green)("The above is the beginning of the process to 'complete the square'")#

#x_("vertex")= (-1/2)xx(color(red)(-4xx15)) = +30#

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So the 60 cm long wire should be cut so that you have 2 lengths of 30 cm each.