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

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