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Yaba asked in Science & MathematicsMathematics · 11 months ago

(1) x > 0. Find the max value of y = x / (x^2 + x + 2)? (2) a > 0, b > 0 and x >= 0. Show that y = x^3 - 3abx + a^3 + b^3 >= 0. How?

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  • 11 months ago

    y = x/(x² + x + 2) ← this is a function, you can obtain the maximum when the derivative is zero

    The function looks like (u/v), so the derivative looks like: [(u'.v) - (v'.u)]/v² → where:

    u = x → u' = 1

    v = x² + x + 2 → v' = 2x + 1

    y' = [(u'.v) - (v'.u)]/v²

    y' = [(x² + x + 2) - (2x + 1).x]/(x² + x + 2)²

    y' = [x² + x + 2 - 2x² - x]/(x² + x + 2)²

    y' = (- x² + 2)/(x² + x + 2)² → then you solve for x the equation: y' = 0

    (- x² + 2)/(x² + x + 2)² = 0 → the denominator cannot be zero

    y = [4√2 - 2] / [16 - 4√2 + 4√2 - 2]

    y = (4√2 - 2) / 14

    y = (2√2 - 1)/7 ← this is the maximum

    • Yaba11 months agoReport

      I got it. Thanks!

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  • 11 months ago

    To show you how far I'll go to avoid the quotient rule... :^)

    First I notice that x^2 + x + 2 > 0 always (negative discriminant and positive at x=0). That means that y>0 for all x>0 and maximizing y is the same as minimizing 1/y.

    1/y = (x^2 + x + 2)/x = x + 1 + 2/x

    (1/y)' = 1 - 2/x^2

    Solving for (1/y)' = 0 gets x^2 = 2 and only the positive square root of 2 fits the x>0 restriction.

    That's problem 1, the easy way.

    For problem 2, find the minimum value for fixed a,b and variable x.

    y'(x) = 3x^2 - 3ab = 0

    x^2 = ab

    x = sqrt(ab)

    The minimum y value on [0, oo) is:

    y(sqrt(ab)) = (ab)^(3/2) - 3ab*sqrt(ab) + a^3 + b^3

    = (ab)^(3/2) - 3*(ab)^(3/2) + a^3 + b^3

    = a^3 - 2*a^(3/2)*b^(3/2) + b^3

    That's the square of [a^(3/2) - b^(3/2)], and a square of a real number is never is never negative. The minimum value of y(x) on [0,oo) can't be negative, y(x) >= 0 for all x>=0.

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