Find two numbers whose sum is 42 and whose product will be at the largest possible.
x + y = 42
y = 42-x
x(42-x)
42x-x^2
derivative 42 -2x =0
x=21 y =21
Find two numbers whose sum is 42 and whose product will be at the largest possible.
When the sum of two numbers is constant, then product of those two numbers is maximum, when those two numbers are as close to each other. As 21 + 21 = 42, so answer will be 21 × 21 = 441.
Oh, I've got just the answer for you! The two numbers you're looking for are 21 and 21. Why? Well, when it comes to getting the largest possible product, you want both numbers to be equal. That way, their product will be maximized. And since 21 + 21 is indeed 42, it's a win-win situation!
To find two numbers whose sum is 42 and whose product will be at the largest possible, we can use the concept of maximizing the product of two numbers given their sum.
Let's assume the two numbers are x and y. According to the problem, we need to find x and y such that their sum is 42.
We can express this as an equation:
x + y = 42
To find the largest possible product, we need to consider that the numbers are positive. Why? Because any negative numbers would reduce the product.
Now, to maximize the product, we need to allocate the largest value to one of the numbers (x or y). We can achieve this by distributing the sum as evenly as possible between the two numbers. In other words, we should choose x and y to be as close to each other as possible.
To find the two numbers, we divide the sum equally between them:
x = 42 / 2 = 21
y = 42 / 2 = 21
So, the two numbers whose sum is 42 and whose product will be the largest possible are 21 and 21.