Two positive numbers have the property that their product is 2 and their sum is as small as possible. Find their sum.
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
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
as usual, for a given perimeter (sum), a square has maximum area (product).
To find the sum of two positive numbers with a product of 2 and a minimum sum, we can start by expressing the two numbers as x and y.
Given that the product of the two numbers is 2, we can write the equation:
x * y = 2
To find the sum of the two numbers, we need to minimize the sum, which means minimizing x + y.
To simplify the problem, let's express one variable in terms of the other variable. For example, we can solve the equation for x:
x = 2 / y
Now, substitute this expression for x into the sum:
(2 / y) + y
To minimize this sum, we can find the value of y that makes its derivative equal to zero. So, let's differentiate the sum with respect to y:
d/dy [(2 / y) + y] = -2/y^2 + 1
Setting this derivative equal to zero:
-2/y^2 + 1 = 0
Rearranging this equation:
-2 = -y^2
Dividing both sides by -1:
2 = y^2
Taking the square root of both sides:
y = ± √2
Since we are looking for positive values, we take the positive square root:
y = √2
Now, substitute this value of y back into the expression for x:
x = 2 / y
x = 2 / √2
x = √2
So, the two numbers that satisfy the conditions are x = √2 and y = √2.
Finally, calculate the sum of these two numbers:
sum = x + y
sum = √2 + √2
sum = 2√2
Therefore, the sum of the two positive numbers is 2√2.