find absolute minimum at (x,y)

f(x) = 2 (square root x) (x − 1); x ≥ 0

Question

find absolute minimum at (x,y)

f(x) = 2 (square root x) (x − 1); x ≥ 0

carson889 · Answer

2*x^(1/2) * (x-1) = 2*x^(3/2) - 2*x^(1/2)
Take derivative: (6/2)(x^(1/2)) - x^(-1/2)
Set equal to zero to find critical point: (6/2)(x^(1/2)) - x^(-1/2) = 0
(6/2)(x^(1/2)) = x^(-1/2)
$$\frac{ 6\sqrt{x} }{ 2 }=\frac{ 1 }{ \sqrt{x} }$$
Multiply both sides by sqrt(x):
$$\frac{ 6x }{ 2 } = 1$$
$$6x = 2$$
$$x = \frac{ 1 }{ 3 }$$
Plug x into original equation to get: 
$$\frac{ -4\sqrt{1/3} }{ 3 }$$

anonymous · Answer

@carson889 i just plugged that into my assignment and it stated it was incorrect. did you write down the problem correctly?

anonymous · Answer

$$2\sqrt{x}(x-1) ; x \ge0$$

anonymous · Answer

nvm i got it. thanks for your help

carson889 · Answer

Your welcome. And glad it worked out.