7.3.5 The Wire-Cutting Problem

Question 1

note

Answer 1

• Values that make the derivative of a function equal to 0 or undefined are candidates for maxima and minima of the function.
• The wire cutting problem involves minimizing the sum of the areas of a square and a circle formed from a fixed length of material.
• For a given perimeter, a circle encloses more area than a square does.

Question 2

note

Answer 2

• In this problem you are asked to find where a wire
  should be cut so that the area made by the resulting
  two pieces is minimized.
• You know the length of the wire, the formulas for the
  area and perimeter of the square, and the formulas
  for the area and circumference of the circle.
• Since the perimeter plus the circumference equals
  the total length of the wire, you can relate the area
  to the length of the wire by expressing the perimeter
  (or sides of the square) in terms of the
  circumference. This will enable you to express the
  area in terms of a single variable, namely c.
• Differentiate the area to find the minimum
  candidate. Do not forget to check that the candidate
  is indeed a minimum.
• Knowing the circumference tells you where to cut
  the wire.
• Circles pack area more efficiently than squares. So
  if you want to minimize the sum of the two areas, it
  makes sense that you would want the square to be
  bigger.

Question 3

A 16 inch wire is cut in two and shaped into two squares. What is the minimum possible sum of the two areas?

Answer 3

Asum = 8 in^2

Question 4

A 12 inch wire is being cut into two pieces which are then shaped into a circle and a regular octagon of maximum area. What is the sum of the perimeters?

Answer 4

12 inches

Question 5

A 24 inch wire is cut in two and shaped into a square and a regular octagon. What is the minimum possible sum of the two areas?AOctagon=(2+2√2)s^2

Answer 5

Asum  = 19.689in^2