Find two positive numbers where the sum of the first and twice the second is 20 and the product is a maximum.?

Method 1 :

Let n be the second positive number.
Then, the first positive number = 20 - 2n

The product
= n * (20 - 2n)
= 20n - 2n²
= -2(n² - 10n)
= -2(n² - 10n + 5²) + 2 * 5²
= -2(n - 5)² + 50

For any real value of n, -2(n - 5)² ≤ 0
Then, the product = -2(n - 5)² + 50 ≤ 50
The maximum product is 50 when n = 5

Hence, the two positive numbers are 10 and 5.


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Method 2 :

Let m be the first positive number, and n be the second.

m + 2n = 20
Then, m = 20 - 2n

The product
= m * n
= (20 - 2n) * n
= 20n - 2n²
= -2(n² - 10n)
= -2(n² - 10n + 5²) + 2 * 5²
= -2(n - 5)² + 50

For any real value of n, -2(n - 5)² ≤ 0
Then, the product = -2(n - 5)² + 50 ≤ 50
The maximum product is 50 when n = 5

Hence, the two positive numbers are 10 and 5.

x + 2y = 20
x = 6, y = 7
The numbers are 6 and 7.

Let the numbers be m and n.
We have m + 2n = 20, and mn = maximum.
m = 20 - 2n, so mn = (20 - 2n)(n) = 20n - 2n².

Take the first derivative:
y = -2n² + 20n, so
y' = -4n + 20.
Set this equal to zero, to find the maximum:
-4n + 20 = 0, or
4n = 20, so
n = 5.

m = 20 - 2n = 20 - 19 = 10.
check:
m + 2n = 10 + 2(5) = 20.
mn = 50 = max.