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

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

Answer 6

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….

Answer 7

1 & 39