AboutAbe Mantell Expertise Hello, I am a college professor of mathematics and regularly teach all levels from elementary mathematics through differential equations, and would be happy to assist anyone with such questions!
Experience Over 15 years teaching at the college level.
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
