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    Max values of a function in a given interval

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    Find the maximum of f(x) = x^4 = 6x^3h + 11x^2h^2 - 6xh^3 at [0,3h]

    © BrainMass Inc. brainmass.com December 24, 2021, 11:05 pm ad1c9bdddf
    https://brainmass.com/math/optimization/max-values-function-interval-536155

    SOLUTION This solution is FREE courtesy of BrainMass!

    f(x) = x^4 - 6 x^3 h + 11x^2 h^2 - 6 x h^3
    can be factorized as:
    f(x) = (x - 0) * (x - h) * (x - 2h) * (x - 3h)

    Hence, f(x) intersects with X axis at points x =0; x = h; x = 2h & x = 3h

    As, in f(x), coefficient of x^4 is +ve, therefore, the function f(x) is monotonically increasing for x > 3h

    After intersection with X axis at x = 3h, function will have a minima in (2h, 3h), followed by a maxima in (h, 2h), and again a minima between (0, h).

    To find maximum or minimum value, differentiate f(x) with respect to x, and equate to zero
    f'(x) = 4 x^3 - 18 x^2 h + 22x h^2 - 6 h^3 = 0
    => x = 3/2h, (3/2 + 1/2 sqrt(5)) h, (3/2 - 1/2 sqrt(5))h
    x = 3h/2 = 1.5 h (between h and 2h)

    x = (3h/2 + sqrt(5)/2)h == 2.61 h (between 2h and 3h)

    x = (3h/2 - sqrt(5)/2)h == 0.38 h (between 0 and h)

    As per the above explanation, the maximum value of f(x) is between h and 2h, i.e., at x = 3h/2

    To confirm mathematically, differentiate f(x) second time,
    f''(x) = 12 x^2 - 36 x h + 22 h^2

    For x = 3h/2
    f''(x=3h/2) = 12*(3h/2)^2 - 36*(3h/2)*h + 22h^2
    = (27 - 54 +22) h^2
    = -5 h^2 (-ve sign indicates maxima)
    Hence, f(x) is maximum in [0,3h] at x = 3h/2
    f_max = f(x =3h/2)
    = (3h/2)^4 - 6 (3h/2)^3 h + 11 (3h/2)^2 h^2 - 6 (3h/2) h^3
    => f_max = (9/16) h^4
    Hence, maximum value of f(x) in interval [0,3h] is (9/16)h^4

    This content was COPIED from BrainMass.com - View the original, and get the already-completed solution here!

    © BrainMass Inc. brainmass.com December 24, 2021, 11:05 pm ad1c9bdddf>
    https://brainmass.com/math/optimization/max-values-function-interval-536155

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