The two numbers are 1 and 1.
Let the non-negative number be x and its reciprocal be y = 1/x
The sum of the two numbers is f(x,y) = x + y
= x + 1/x
= f(x)
For f(x) to be minimum, we differentiate it with respect to x and equate it to zero to find the value of x at which irt is minimum.
So, df(x)/dx = d(x + 1/x)/dx
= dx/dx + d(1/x)/dx
= 1 - 1/x²
Equating it to zero, we have
1 - 1/x² = 0
1 = 1/x²
x² = 1
Taking square-root of both sides, we have
x = ±√1
x = ±1
Since x is non-negative x = + 1
We differentiate f(x) again to determine if this value gives a minimum for f(x).
So, d²f(x)/dx² = d(1 - 1/x²)/dx
= d1/dx - d(1/x²)/dx
= 0 - × (-2/x³)
= 2/x³
Substituting x = 1 into the equation,
d²f(x)/dx² = 2/x³
= 2/1³
= 2 > 0.
So x = 1 is a minimum point for f(x)
Since x = 1 and y = 1/x, y = 1/1 = 1
So, the two numbers are 1 and 1.
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