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Finding the stationary point of a function of the form $\\frac{a \\ln(x)}{x}$ and determining its nature.

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Given the function \\[ \\simplify{y={a}ln(x)/x} ,\\] find its stationary point and determine its nature.

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To find the stationary point of the function, we must solve $\\tfrac{dy}{dx}=0$ for $x$. For the function $\\simplify{y={a}ln(x)/x}$, using the quotient rule,

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\\[ \\frac{dy}{dx}= \\frac{\\simplify{{a} - {a}ln(x)}}{x^2}.\\]

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Setting $\\frac{dy}{dx}=0$ and solving for $x$:

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\\[ \\begin{split} \\frac{\\simplify{{a}-{a}ln(x)}}{x^2} = 0 \\\\ \\implies \\simplify{{a}-{a}ln(x)} = 0 \\\\ \\implies 1-\\ln(x) =0 \\end{split}\\]

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Therefore,

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\\[ \\begin{split} \\ln(x)&\\,=1 \\\\ \\implies x&\\,=e \\\\ &\\,=2.72 \\text{  (2 d.p.)}\\end{split} \\]

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The corresponding $y$-coordinate can be found by plugging $x=e$ into the initial equation:

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\\[ \\begin{split} y &\\,= \\frac{\\simplify{{a}ln(e)}}{e} \\\\ &\\,= \\frac{\\var{a}}{e} \\\\ &\\,= \\var{precround(a/e ,2)} \\text{ (2 d.p.)} \\end{split} \\]

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Hence, the function has one stationary point at \\[ (x,y) = (\\var{solx},\\var{soly}) .\\]

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To determine the nature of the stationary point we want to calculate the second derivative of the function and then evaluate it for the $x$-coordinate of the stationary point.

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Recall:

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To calculate $\\tfrac{d^2y}{dx^2}$, we want to differentiate $\\tfrac{dy}{dx}$ again with respect to $x$:

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\\[ \\frac{dy}{dx} = \\frac{\\simplify{{a} - {a}ln(x)}}{x^2}\\]

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\\[ \\begin{split} \\implies \\frac{d^2y}{dx^2} &\\,=\\frac{x^2 \\left(\\frac{\\var{-a}}{x}\\right)-2x(\\var{a}-\\var{a}\\ln(x))}{x^4} \\\\\\\\ &\\,=\\frac{\\var{-3a}x+\\var{2a}x \\ln(x)}{x^4} . \\end{split}\\]

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When $x=e$, 

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\\[ \\begin{split} \\frac{d^2y}{dx^2} &\\,= \\frac{\\var{-3a}e+\\var{2a}e}{e^4} \\\\ &\\,= \\frac{\\var{-a}}{e^3}  \\\\ &\\,\\approx\\var{precround(-a/(e^3),2)}\\,. \\end{split} \\]

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Therefore, $\\tfrac{d^2y}{dx^2}<0$ and the stationary point is a maximum.

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There is a stationary point at ([[0]],[[1]]) .

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The stationary point is a: 

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