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(x38) = x^2  38x
dP/dx = 2x  38 = 0
x = 19

361 i think..
xy=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