Find three positive numbers whose sum is 231 and whose product is a maximum.
a cube has maximum volume for a given surface area, so you want numbers which are all about equal.
231/3 = 77
77*77*77 = 456,533
You can check a few alternatives.
77*76*78 = 456,456
and so on
To find three positive numbers whose sum is 231 and whose product is a maximum, we can use the concept of optimizing functions. Let's assume the three numbers as x, y, and z.
Since their sum is given as 231, we have the equation:
x + y + z = 231
To maximize the product, we'll express z in terms of x and y. Rearranging the equation, we get:
z = 231 - x - y
Now, we can express the product P in terms of x and y:
P = x * y * z
Substituting the value of z, we have:
P = x * y * (231 - x - y)
To find the maximum value of the product P, we need to find the critical points. To do that, we'll take the partial derivatives of P with respect to x and y and set them equal to zero.
dP/dx = y * (231 - 2x - y) = 0
dP/dy = x * (231 - 2y - x) = 0
From the above equations, we can determine the critical points where the product might be maximized. Solving these equations can be quite complex, so let's use an alternative method.
We will use the concept of symmetry to simplify the problem. If we have three equal numbers whose sum is 231, the product will be maximum. Therefore, we'll divide 231 by 3 to get the equal value, which is approximately 77.
Thus, the three positive numbers are 77, 77, and 77, with a sum of 231 and a maximum product.
Please note that this is just one possible solution; there may be other combinations that yield the same maximum product.