To find two positive numbers where the sum of the first number and twice the second equals 56, and whose product is maximized, we can use algebraic methods coupled with calculus.
Let the first number be x and the second number be y. According to the problem, we have the equation:
x + 2y = 56
We want to maximize the product P = xy. We can express y in terms of x using the first equation:
y = (56 – x) / 2
Substituting this expression for y into the product equation:
P = x * ((56 – x) / 2) = (56x – x²) / 2
To maximize P, we can find its derivative and set it to zero:
P’ = (56 – 2x) / 2
Setting the derivative equal to zero:
56 – 2x = 0
This simplifies to:
x = 28
Substituting back to find y:
y = (56 – 28) / 2 = 14
Thus, the two positive numbers are 28 and 14.
Finally, we can check that these numbers satisfy the original condition:
28 + 2(14) = 28 + 28 = 56 (Condition satisfied)
The product is:
P = 28 * 14 = 392
Therefore, the two positive numbers we seek are 28 and 14, which gives a maximum product of 392.