Find two numbers differing by 38 whose product is as small as possible?

Find two numbers differing by 38 whose product is as small as possible.?

How do i start this

7 Answers

  • Let the numbers are A & a+38

    P=A*(A+38)=A^2 +38A

    P' = 2A +38

    for minimum p' =0

    2A +38 =0

    2A =-38

    A =-38/2 =-19

    A+38 =-19+38 =19

    The numbers are 19 and -19

  • let's say one number is : x and the other one is : x + 38

    P = x*(x+38) = x^2 + 38 x

    P' = 2 x + 38

    P'' = 2

    P' = 0 ==>> 2 x + 38 = 0 => x = -38 / 2 = -19

    So :

    x = -19 , (x + 38) = -19 + 38 = 19

    the two numbers are -19 and 19.

  • What numbers are you looking for? Only Positive? It will not work.

    If x and x+ 38 are the numbers, you want P = x(x+38) to be minimum.

    So write P = x^2 + 38x = x^2 + 38x + 361 - 361 = ( x+19)^2 - 361. This is minimum if x = -19. So -19 and 19 are the numbers. Only Positive will not work.

  • How about 0 and 38

  • The answer will be -(19^2) = -361. Here's how you find it:

    P = x(x-38) = x^2 - 38x

    dP/dx = 2x - 38 = 0

    x = 19

  • 361 i think..

    x-y=38

    xy=f(x)

    y(38+y) shud b minimum,

    let y(38+y)=f(y)

    differentiate equate it to zero...get the value of y...

    and x and the product....

  • 1 & 39

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