To find three positive numbers whose sum is 100 and whose product is maximized, we can use the method of Lagrange multipliers or simply apply some algebraic insights.
Let’s denote the three numbers as x, y, and z. According to the problem, we have:
- x + y + z = 100
To maximize the product P = xyz, we can express z in terms of x and y: z = 100 – x – y. So, we can rewrite the product as:
P = xy(100 – x – y) = 100xy – x^2y – xy^2
To find the critical points, we would typically take the partial derivatives with respect to x and y, set them to zero, and solve. However, we can also utilize the symmetry of the situation.
By the AM-GM inequality (Arithmetic Mean – Geometric Mean Inequality), for three positive numbers, the product is maximized when all three numbers are equal. Thus, we set:
x = y = z
Substituting this into the sum equation:
3x = 100
Solving for x, we get:
x = 100/3 ≈ 33.33
Therefore, the three numbers are:
x ≈ 33.33, y ≈ 33.33, z ≈ 33.33
In conclusion, the three positive numbers that sum up to 100 and maximize the product are approximately 33.33, 33.33, and 33.33.