What is the largest area it can enclose?
A hard-level "Area" problem.The same perimeter can enclose wildly different areas. "With a fixed sum, the product peaks where the two are equal" shows up in fences, fields and boxes alike.
GivenA rope 20 m long is used to fence off a rectangle. The sides are whole numbers of metres.
Hints
- The perimeter is (height + width) × 2, so height + width = 20 ÷ 2 = 10 m. However the rectangle is reshaped, that sum never changes. That is the starting point.
- List every pair adding to 10: 1 and 9, 2 and 8, 3 and 7, 4 and 6, 5 and 5. Here they are, stacked with their bottom-left corners together.
- Notice something: the top-right corners fall on a perfect straight line. Add one to the height and the width loses one — the sum is fixed, so it has to happen.
- Their areas run 9, 16, 21, 24, 25. The thinner it is the less it holds; the closer to a square, the more. Same perimeter, yet the area swings from 9 all the way to 25.
- The best is the 5 m × 5 m square: 25 m². When the sum is fixed, the product is largest where the two are equal. The same shape of argument turns up again and again.
- One last thing. What if the shape need not be a rectangle? The same 20 m of rope bent into a circle encloses about 31.8 m² — more than the square. For a given perimeter, the circle holds the most. That is the last move of these hundred problems.
Answer 25m²
More "Area" problems
- Area of the inner triangle?
- Area of triangle DOA?
- Area of quadrilateral ABCD?
- Area of the overlap?
- Total area of the two crescents?
- Area swept by the square?