So, fun fact from calculus (though you can easily prove this with basic algebra as I do below):
- You want to compare two products: in this case 6x9 and 7x8.
- And in each product, if you add the two numbers together, you get the same result. In this case, 6+9 = 7+8.
Then the product will be larger for the pair of numbers that are closer together. So 7x8 > 6x9. That might help you remember which is 56 and which is 54.
You typically see this in a word problem where you are given a fixed amount of fence and you have to enclose the largest rectangular area. The answer is to use a square area (two sides being equal). If the problem has constraints that prevent the sides from being equal, then you pick the length and width to be as close to each other as possible.
In case you want to transfer the geometric intuition to an algebraic proof: If the sum of the two side lengths is 2m, then the two side lengths can be written as (m+n) and (m-n) for some positive n. If you multiply the two, you get (m+n)(m-n) = m²-n². To maximize the product, you need n to be as close to 0 as possible - i.e. for both sides to be as close to each other as possible.
- You want to compare two products: in this case 6x9 and 7x8.
- And in each product, if you add the two numbers together, you get the same result. In this case, 6+9 = 7+8.
Then the product will be larger for the pair of numbers that are closer together. So 7x8 > 6x9. That might help you remember which is 56 and which is 54.
You typically see this in a word problem where you are given a fixed amount of fence and you have to enclose the largest rectangular area. The answer is to use a square area (two sides being equal). If the problem has constraints that prevent the sides from being equal, then you pick the length and width to be as close to each other as possible.
In case you want to transfer the geometric intuition to an algebraic proof: If the sum of the two side lengths is 2m, then the two side lengths can be written as (m+n) and (m-n) for some positive n. If you multiply the two, you get (m+n)(m-n) = m²-n². To maximize the product, you need n to be as close to 0 as possible - i.e. for both sides to be as close to each other as possible.