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
