Respuesta :
We are given equations as
[tex]xy=12[/tex]
Firstly, we will solve for y
[tex]y=\frac{12}{x}[/tex]
now, we are given sum as
[tex]S=2x+y[/tex]
now, we can plug back y
[tex]S=2x+\frac{12}{x}[/tex]
Since, we have to minimize it
so, we will find it's derivative
[tex]S'=2-\frac{12}{x^2}[/tex]
now, we can set it to 0
and then we can solve for x
[tex]S'=2-\frac{12}{x^2}=0[/tex]
[tex]x=\sqrt{6},\:x=-\sqrt{6}[/tex]
Since , both numbers are positive
so, we will only consider positive value
[tex]x=\sqrt{6}[/tex]
now, we can find y-value
[tex]y=\frac{12}{\sqrt{6}}[/tex]
[tex]y=2\sqrt{6}[/tex]
So, two positive numbers are
[tex]x=\sqrt{6}[/tex]
[tex]y=2\sqrt{6}[/tex]............Answer
Answer:
The positive numbers are [tex]$x=\sqrt{6}, \quad\\ y=2 \sqrt{6}$[/tex].
Explanation:
[tex]$S^{\prime}(x)=0$[/tex]
The equation is derived as,
[tex]$\Rightarrow \frac{d}{d x}\left(2 x+12 x^{-1}\right)=0$[/tex]
[tex]$\Rightarrow 2-12 x^{-2}=0$[/tex]
[tex]$\Rightarrow x^{2}=6$[/tex]
Further simplified as,
[tex]$x=\pm \sqrt{6}$[/tex]
[tex]$x=\sqrt{6}$[/tex]
[tex]$y=\frac{12}{\sqrt{6}}\\=2 \sqrt{6}$[/tex]
Since we have only one local minimum on our open interval [tex]$\backslash x \in(0, \infty), \backslash$[/tex] it's the absolute minimum.
Learn about refer bonds refer:
- https://brainly.com/question/19139062.
