Find a positive number x such that the sum of 16x and 1/x is as small as
Does this require an open or closed interval?
- 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.
- BryceLv 74 months ago
y'= 16 - 1/x²= 0