Find positive numbers x and y satisfying the equation xy=12 such that the sum 2x+y is as small as possible.

Question

Find positive numbers x and y satisfying the equation xy=12 such that the sum 2x+y is as small as possible.

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    0
    2021-09-10T06:32:56+00:00

    We are given equations as

    xy=12

    Firstly, we will solve for y

    y=\frac{12}{x}

    now, we are given sum as

    S=2x+y

    now, we can plug back y

    S=2x+\frac{12}{x}

    Since, we have to minimize it

    so, we will find it’s derivative

    S'=2-\frac{12}{x^2}

    now, we can set it to 0

    and then we can solve for x

    S'=2-\frac{12}{x^2}=0

    x=\sqrt{6},\:x=-\sqrt{6}

    Since , both numbers are positive

    so, we will only consider positive value

    x=\sqrt{6}

    now, we can find y-value

    y=\frac{12}{\sqrt{6}}

    y=2\sqrt{6}

    So, two positive numbers are

    x=\sqrt{6}

    y=2\sqrt{6}…………Answer

    0
    2021-09-10T06:33:23+00:00

    Answer:

    The positive numbers are $x=\sqrt{6}, \quad\\ y=2 \sqrt{6}$.

    Explanation:

    $S^{\prime}(x)=0$

    The equation is derived as,

    $\Rightarrow \frac{d}{d x}\left(2 x+12 x^{-1}\right)=0$

    $\Rightarrow 2-12 x^{-2}=0$

    $\Rightarrow x^{2}=6$

    Further simplified as,

    $x=\pm \sqrt{6}$

    $x=\sqrt{6}$

    $y=\frac{12}{\sqrt{6}}\\=2 \sqrt{6}$

    Since we have only one local minimum on our open interval $\backslash x \in(0, \infty), \backslash$ it’s the absolute minimum.

    Learn about refer bonds refer:

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