Expert: Abe Mantell - 6/7/2009

Question
A strip of wire of length 28cm is cut into two pieces.One piece is bent to form a square of side x cm,and the other piece is bent to form a rectangle of width 3 cm.

a)Show that the length of the other side of the rectangle is given by (11-2x)cm.
b)Deduce that the total combined area of the square and the rectangle is (x²-6x+33)cm².
c)Prove that the minimum total area which can be enclosed in this way is 24cm².

Answer
If the square has sides of length x, then the total perimeter is 4x.
Thus, 28-4x is left over for the rectangle. If the rectangle has 2 sides
that are 3, then 28-4x-3-3 is left over for the other two sides.
28-4x-3-3=22-4x, so each of the remaining sides must be (22-4x)/2=11-2x

Area of the square is x^2, Area of the rectangle is 3*(11-2x)=33-6x
Thus, the combined area is x^2 + 33 - 6x = x^2 - 6x + 33

The minimum of this quadratic occurs for x=-b/2a = -(-6)/(2*1)=3

So, for a minimum total area x=3 ==> 3^2 - 6*3 +33 = 6-18+33 = 24 sq. cm

Abe