a b = 2 ... a = 2 / b
s = a + b = 2/b + b
ds = db (-2/b² + 1)
ds/db = -2/b² + 1 = 0
b = √2
s = 2√2
Two positive numbers have the property that their product is 2 and their sum is as small as possible. Find their sum.
3 answers
Let one of them be x, and the other one y
xy = 2
y = 2/x
sum = x+y = x + 2/x
d(sum)/dx = 1 - 2/x^2 = 0 for a min sum
1 = 2/x^2
x^2 = 2
x = √2 , then y = 2/√2 = √2
their minimum sum = x+y
= √2 + √2 = 2√2
xy = 2
y = 2/x
sum = x+y = x + 2/x
d(sum)/dx = 1 - 2/x^2 = 0 for a min sum
1 = 2/x^2
x^2 = 2
x = √2 , then y = 2/√2 = √2
their minimum sum = x+y
= √2 + √2 = 2√2
as usual, for a given perimeter (sum), a square has maximum area (product).