Calculus Optimization problem

Question

Two nonnegative numbers have a sum of 10. What is the least possible sum of their reciprocals?

 

The answer is 2/5

 

Answer 1

Let the numbers be x and y.  So, since x+y=10, y=10-x.

 

Minimize:  f(x) = 1/x + 1/(10-x), where 0<x<10.

 

               f'(x) = -x^-2 + -(10-x)^-2(-1)

 

                      = -1/x² + 1/(10-x)²

 

                      = [-(10-x)² +x²]/[x²(10-x)²]

 

                      = [-(100-20x+x²) +x²]/[x²(10-x)²]

 

                      = (20x-100)/[x²(10-x)²]

 

                      = 0 when x = 5

 

When 0<x<5, f'(x) < 0

When 5<x<10, f'(x) > 0

 

So, f(x) is minimized when x=y=5

 

Minimum sum = f(5) = 1/5 + 1/5 = 2/5