# Math Problem?

Find a positive number x such that the sum of 16x and 1/x is as small as

possible

Does this require an open or closed interval?

### 3 Answers

Relevance

- atsuoLv 64 months ago
x is positive, so x = (0,inf) ← open interval

16x + 1/x

= (4√x - 1/√x)^2 + 8

Clearly (4√x - 1/√x)^2 has its minimum value 0 when

4√x - 1/√x = 0

4x - 1 = 0

x = 1/4

So 16x + 1/x has its minimum value 8 when x = 1/4.

- sepiaLv 74 months ago
Find a positive number x such that

the sum of 16x and 1/x is as small as possible.

y = 16x + 1/x

dy/dx = 16 - 1/x²

dy/dx = 0 when 16 -1/x² = 0 that's when x² = 1/16, i.e. when x = 1/4.

When x = 1/4, the sum of 16 x and 1/x is 8.

That's the minimum value over the open interval x > 0.

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