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Optimization: Find the point P on the graph of the function y=sqrt(x) closest to the point (6,0)?

Find the point P on the graph of the function y=sqrt(x) closest to the point (6,0)

The x coordinate of P is: ______

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  • 11 months ago
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    y = √x ← this is the function, i.e. a curve

    The point P belongs to the curve and closest to the point A (6 ; 0).

    Let's calculate the distance [AP]

    xAP = xP - xA = xP - 6

    yAP = yP - yA = yP

    AP² = xAP² + yAP²

    AP² = (xP - 6)² + yP²

    AP² = xP² - 12xP + 36 + yP² → as the point P on the curve: y = √x → yP = √xP → yP² = xP

    AP² = xP² - 12xP + 36 + xP

    AP² = xP² - 11xP + 36 ← you can see that the distance is a function of xP

    g(x) = xP² - 11xP + 36

    If you want to obtain a minimum for [AP], so you must obtain the minimum for AP².

    You can obtain the minimum of a function when its derivative is zero.

    g'(x) = 2xP - 11 → then you solve the equation: g'(x) = 0

    2xP - 11 = 0

    xP = 11/2 ← this is the abscissa of the point P

    Recall: yP = √xP ← because P is on the curve: y = √x

    yP = √(11/2)

    yP = (√22)/2 ← this is the ordinate of the point P

    → P [11/2 ; (√22)/2]

    To go further, let's calculate the distnce [AP]

    AP² = xP² - 11xP + 36 → we've just seen that: xP = 11/2

    AP² = (11/2)² - 11.(11/2) + 36

    AP² = (121/4) - (242/4) + (144/4)

    AP² = 23/4

    AP = (√23)/2

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  • 11 months ago

    The closest point would be where the tangent to the curve is perpendicular to the connecting line.

    y = √x

    dy/dx = (1/2)x^(-1/2) = 1/(2√x)

    slope at x = x₀ is 1/(2√x₀)

    perpendicular is a slope of –2√x₀

    line perpendicular is

    y = –(2√x₀)x + k

    0 = –(2√x₀)6 + k

    k = –12√x₀

    equation is

    y = –(2√x₀)x – 12√x₀

    we need distance from point x₀,√x₀ to 6,0

    distance between two points

    d = √(Δx² + Δy²)

    d² = (x₀–6)² + (√x₀)²

    d² = x₀² – 12x₀ + 36 + x₀

    d² = x₀² – 11x₀ + 36

    this has a minimum at about x₀ = 5.5

    d = √(x₀² – 11x₀ + 36)

    dd/dx₀ = [(1/2)(x₀² – 11x₀ + 36)^(–1/2)] [ 2x₀ – 11 ] = 0

    [ 2x₀ – 11 ] / √(x₀² – 11x₀ + 36) = 0

    (2x₀ – 11) = 0

    x = 5.5

    y = √5.5

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  • 11 months ago

    let the pt be (x, sqrt(x))

    Let s be the distance

    then s^2 = (x-6)^2 + (sqrt(x) - 0)^2 = (x-6)^2 + x

    d(s^s)/dx = 2(x-6) +x = 0

    or 3x = 12 or x = 4

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