the sum of 2 nonnegative numbers is 30. Find numbers a) if the sum of their squares is a minimum, b) if sum of their squares is a maximum ?

Using calculus,how would you do this?

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  • 4 weeks ago
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    The sum of two non-negative numbers is 30:

    x + y = 30 where x > 0 and y > 0

    Find the numbers if the sum of their square is a minimum:

    We can set up this equation:

    s = x² + y²

    If we solve the first equation for x in terms of y we get:

    x = 30 - y

    Substitute and simplify into the second equation to get:

    s = (30 - y)² + y²

    s = 900 - 60y + y² + y²

    s = 2y² - 60y + 900

    The value of "y" that gives "s" a minimum can be found by solving for the zero of the first derivative:

    ds/dy = 4y - 60

    0 = 4y - 60

    -4y = -60

    y = 15

    Now that we have y, solve for x:

    x = 30 - y

    x = 30 - 15

    x = 15

    Both numbers are 15 to make "s" the minimum value of 450.

    • Puzzling
      Lv 7
      4 weeks agoReport

      Non-negative includes zero. So x ≥ 0 and y ≥ 0. That's important for the maximum sum cases (0 and 30)

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  • Anonymous
    4 weeks ago

    a) 14 + 16 .....450

    b) 29 + 1 .......842

  • 4 weeks ago

    Let x be one of the numbers. (30-x) is the other number.

    The sum of their squares is x^2 +(30-x)^2 

    = x^2 + 900 - 60x + x^2

    = 2(x^2) - 60x + 900

    At both the maximum and the minimum, the slope will be 0 or the point will be at a boundry.

    Slope at x is 4x - 60

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  • 4 weeks ago

    Let x be the first non-negative number (x ≥ 0)

    Let 30 - x be the second non-negative number (30 - x ≥ 0)

    If you check both the inequalities, you see that both numbers are between 0 and 30 inclusive.

    Intuitively, you should be able to figure out the minimum is when both numbers are equal. And the maximum would be when one is 0 and one is 30.

    But let's be more formal and solve it using calculus.

    Write a formula for the sum of their squares.

    f(x) = x² + (30 - x)²

    f(x) = x² + 900 - 60x + x²

    f(x) = 2x² - 60x + 900

    If you graph this (or remember algebra II), you'll find it is an upward facing parabola and you could figure out the vertex is a minimum when x = 15. From this you could get the answer.

    Again, let's forget the easier route and continue using calculus.

    Take the derivative:

    f'(x) = 4x - 60

    Set it to zero:

    f'(x) = 4x - 60 = 0

    4x = 60

    x = 15

    But you don't know if this is a maximum or minimum. You could take the second derivative.

    f"(x) = 4

    Since this is positive that means the curve is concave up and so x = 15 would be a minimum.

    You'd then have to look at the graph from 0 to 30 to see that the maximums are at the endpoints.

    Answer:

    15 and 15 --> minimum

    0 and 30 (or 30 and 0) --> maximum

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