Page 68 - Calc for the Clueless

Given a 68-inch string, divide it into two pieces, one a square and one a rectangle whose length is twice its
width, and find the minimum and maximum possible total area of the figures.

Let one piece be x. The other one will be 68 - x.

If x is the perimeter of the rectangle, then x = 6w. So w = x/6 annnd l = 2w = x/3. The area of the rectangle is
2
x /18.

The perimeter of the square is 68 - x. Since p = 4s, s = p/4 = (68 - x)/4. The area of the square is (68 - x) /16.
2
2
The total area A = x /18 + (68 - x) /16. dA/dx = x /9 -(68 - x)/8 = 0.
2
2
Cross multiplying after bringing the negative term to the other side, we get

2
The area would be 36 /18 + (68 - 36) /16 = 72 + 64 = 136.
2
But there are two other possible values for the max and min: if the string is uncut, the whole string may be a
rectangle or the whole string might be a square. If the string is a square, s = p/4 = 68/4 = 17. A = s = 289. If the
2
string is a rectangle, w = p/6 = 68/6; l = 2w = 68/3.

Note
In some books, the string is cut into a square and an equilateral triangle, or
sometimes———yech! ———a square and a circle (quite messy).

The largest number is the maximum area, when the figure is a square (289). The smallest number is the
minimum area, when the string is cut (36'' for the perimeter of the rectangle, 32" for the square, minimum area =
136).
So far we have limited the problems to the relatively gentle. However, there are some pretty nasty problems in
some books. Let's try four of them.

Example 12—

Joan's house has a window in the shape of a rectangle surmounted with a semicircle. For a given perimeter p,
what are the dimensions of the window if it allows the maximum amount of light?

We first note that maximum light means maximum area. If we are careful and a little clever, the problem is
messy but not too bad.

We note we must maximize the area, which is a rectangle plus half a circle.