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Applied Maximum and Minimum: Calculus?

How should two nonnegative numbers be chosen so that their sum is 1 and the sum of their squares is (a) as large as possible, (b) as small as possible.

3 Answers

  • Anonymous
    1 decade ago
    Favorite Answer

    x + y = 1

    y = 1 - x

    g = x^2 + (1 - x)^2 = x^2 + 1 - 2x + x^2 = 2x^2 - 2x + 1

    g' = 4x - 2 ; 0 = 4x -2 ; 2 = 4x ; x = 1/2 and therefore y = 1/2

    so theres a max or min at x = 1/2

    g'' = 4, so it is concave up therefore x = 1/2 is a minimum

    and because there are no maximum, we know that the highest possible value of g are at its boundaries ( end points )( either x = 0 or x =1). In this case x =0 and x = 1 give the same g so they are both the highest possible value of g

  • 1 decade ago

    let the two numbers be x and y

    x^2 + y^2 = a where a is a constant


    a) x^2 + y^2 =a

    2x + 2y(dy/dx) = 1

    dy/dx = (1-2x)/2y

    when eqn is max or min, dy/dx = 0

    (1-2x)/2y = 0

    1-2x = 0

    x = 0.5

    y = 0.5

    To find the max value, simply look at the graph of the function

    x^2 + y^2 = a

    This is the equation for the graph of a circle with center (0,0)

    when value is max or min, dy/dx = 0, which means the gradient is 0

    The only part of the graph where this happens is on the y-axis ie when x=0 and y = 1 or y=-1. but as your numbers must be positive, y=1

  • Anonymous
    1 decade ago

    Let the two numbers be a and b

    then a+b=1


    Sume of their squares - S=a^2+b^2

    subs in for b




    for max and min put dS/da=0


    a=1/2, b=1/2 these values give minimum

    maximum is when one is 1 and other is 0.

    If you plot the graph f(a)=2a^2-2a+1 you will see that the maximum values are not turning points so will not be calculated by this method.

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