My wife wanted to bake something and needed a 8x8inch pan. Don’t have one but did have a 10 x 6 inch. (Same size, Different shape.) If you measure the four sides of each pan you come up with 32 inches, but if you figure out the square inches of each pan, there is a 4 square inch difference. (8x8 =64 sq. inches, 10 x 6 = 60 sq. inches. Can anybody explain!
If you want to maximize an area enclosed by a particular perimeter and are constrained to a rectangular shape, the max area will always be a square.
If you can go freeform, the max area will be a circle.
Someone will be along shortly to show the math. Basically, in your example, consider a hypothetical pan that is 1 inch by 15 inches. It will have a fairly minimal surface area of 15 square inches. But 8 x 8 gives you the max (rectangular) area of 64 square inches.
Yup. If you want to max area, a circle is your best bet. If it has to be rectangular, a square is best.
Multiplying different numbers often yields different products. This is true both in the world of mathematics and that of baking pans.
And if you want to minimize the area enclosed by a given length of sides, take one pair of sides down to zero. Then you’ll have two long sides touching each other with nothing inside. The point is, the linear measure of the sides will place a limit on the maximum area, but there’s no principle that states that the area should be the same for a given total side length in various configurations. Why should there be?
As a visual aid, take 32" of string tied together to form a loop. You can circumscribe the 8x8 pan, the 10x6 pan, a 15x1 pan, and a (theoretical) 16x0 pan, as well as any number of intermediate layouts. Play around with this string – not just making rectangles, make a circle, ovals, odd squishy shapes, etc. – and you’ll see that while the length of the string dictates a maximum area, it in no way dictates a constant area.
It’s the same reason a soap bubble forms a sphere. There’s a certain amount of air in the bubble, and it can’t get out; but the actual soap is stretched like a rubber band, it would like to shrink back to a small size. So what you get is the smallest surface that can hold a certain volume; a sphere.
In two dimensions, the smallest perimeter to enclose a given area is a circle. And if you insist on having right-angle corners, it’s a square.
If you need a better visual, this Dr. Mathreply may be useful.
If you’re allowed to use one wall of the oven in lieu of pan rim, a semicircle is best. This is known as Dido’s Problem and was allegedly discovered by Aeneas’ lover, the Queen of Carthage.
That the sphere maximizes volume was not proved rigorously until the 19th century. Yet soap bubbles and other things solve such math problems “effortlessly.” Laplace commented on this: “Nature laughs at the difficulties of integration.”
The pans are not the same size. They have the same perimeter, but not the same area. The same perimeter is kind of neat but sort of a coincidence. Would you say that a pan that was 15 inches long but one inch wide (perimeter = 32, area = 15 in²) was the same size? In your example, the two are close, which leads to confusion. I think the more extreme examples make it clearer.
The problem is the word “size.” It is too simplistic to describe a comparison between multi-dimensional entities.
Perimeter is a linear measure. Area is a square measure. One does not depend on the other, as they’re not of comparable units.
Well, this is glib. The lack of dependence in this case doesn’t follow directly from the difference in units. Certainly, there is a quadratic relationship between the two, for a fixed “shape”: e.g., two circles have the same perimeter just in case they have the same area, two squares have the same perimeter just in case they have the same area, two rectangles with sides in a 3:1 ratio have the same perimeter just in case they have the same area, and so on
Think of it this way: Take an empty can. It’s got some volume, right? Now put it on its side on the floor and stomp on it. You didn’t remove any of the metal of the can, so it’s still got the same circumference, but it’s obviously got a lot less volume now, since it’s flattened.
Well, your 10x6 pan is more flattened than your 8x8 pan, so it holds less, too.
With the (reasonable? unreasonable? I have no idea) assumption that stomping a can doesn’t compress or stretch any of the metal in it.
Well, you could always consider a cardboard box. They are generally sold flat. So if you have a box of given dimensions with the top and bottom cut off, you have a perimeter of a 2w + 2l. As you collapse it into a parallelogram, the area is reduced. There is no compression or stretching.
There will be some, there always is. It’s likely to be negligible, and certainly nowhere near enough to account for the volume change. It’s a better assumption than pretty much anything you’ll see in an undergrad engineering course.
For those familiar with calculus, the maximum area is calculated as follows:
(1) Area = Length x Width, or A = L x W for short
(2) We also know that the perimeter is 32, so 32 = 2 x L + 2 x W. This can be rearranged to give W = 16 - L.
(3) Substitute this back into the equation for A, and we get A = L x (16 - L), or A = 16L - (L x L).
(4) To find the maximum value for A, take the first derivative of the formula for A (with respect to L) and set it to zero. This gives us 0 = 16 - 2 x L.
(5) Rearrange to get L = 8. Substitute back into W = 16 - L, and we see that W also equals 8.
Try this with different perimeters, and the answer will always be L = W = 1/4 of the perimeter.
Who says calculus isn’t useful?