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?

Question

Hazel Anglin · Accepted Answer

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

Kennedy Adams · Answer

undefined To minimize the total area of the two squares, the wire should be cut into two equal pieces of 30 cm each. Then each piece should be formed into a square with side length of 15 cm.