Find two positive real numbers such that they add to 52 and their product is as large as possible.?

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  • 2 months ago

    TWO makes me think 25 and 27 (if they must be whole numbers). Otherwise 26 (minus a tiny fraction) and 26 (plus the same tiny fraction). [Calculus gives the proof that a number squared always results in the largest number (area, etc).

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  • Philip
    Lv 6
    2 months ago

    Let numbers be x, 52-x.

    Their product = f(x) = 52x - x^2.= = x(52-x).

    f'(x) = 52 - 2x.

    f''(x) = -2.

    An extremum occurs at f'(x) = 0, ie., at x = 26.

    Since f''(x) < 0, any extremum reached will be

    a maximum.

    f(26) = 26(52-26) = 26^2 = 4*13^2 = 4*(170-1)

    = 676. Both pos. real numbers = 26 and f(x) reaches its maximum at (26,676).

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  • sepia
    Lv 7
    2 months ago

    Find two positive real numbers such that they add to 52 

    and their product is as large as possible.

    Let the two positive real numbers be x and y

    x + y = 52

    xy = 26^2

    (52 - y)y = 26^2

    y^2 - 52y + 26^2 = 0

    y = 26, x = 26

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  • 2 months ago

    Let x and y = your two positive real numbers.

    The sum is 52, so:

    x + y = 52

    Let's solve for y in terms of x:

    y = 52 - x

    You want the largest product possible, so we can set up this function using two unknowns:

    f(x, y) = xy

    We have y in terms of x so we can substitute to make the function have one unknown:

    f(x) = x(52 - x)

    f(x) = 52x - x²

    The "x" that gives the maximum can be found by solving for the zero of the first derivative:

    f'(x) = 52 - 2x

    0 = 52 - 2x

    2x = 52

    x = 26

    Now that we have x, we can solve for y:

    y = 52 - x

    y = 52 - 26

    y = 26

    The largest product you can get from two real numbers with the sum of 52 is:

    xy

    26(26)

    676

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  • 2 months ago

    26 and 26. Note that

    x^2 > (x-1)(x+1) > (x-2)(x+2) > (x-3)(x+3) > and so forth.

    So you could put ANY even number into your question,

    and the answer would always be, split it in half.

    • Dixon
      Lv 7
      2 months agoReport

      Nice, there's always one more trick using the difference of two squares :-)

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  • Dixon
    Lv 7
    2 months ago

    If there is a maximum (and by inspection, there is) then by symmetry it must be when the two numbers are equal., ie 26 and 26.

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