Find two numbers whose difference is 100 and whose product is a minimum.

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i1ziZ · Accepted Answer

It is given that the difference of two numbers is 100.
x- smaller
y-larger
$y - x = 100$
$y = 100 + x$
We write a function that represents the minimum product of two numbers.
$f (x, y) = x y$
Substitute $y = 100 + x$ into $f (x, y) = x y$ .
$f (x) = x (100 + x)$
$= 100 x + x^{2}$
Now, we find the first derivative of the function $f (x) = 100 x + x^{2}$ .
$f^{'} (x) = {(100 x + x^{2})}^{'}$
$= {(100 x)}^{'} + {(x^{2})}^{'}$
$= 100 + 2 x$
Find the critical points. Solve the equation $f^{'} (x) = 0$ .
$f^{'} (x) = 100 + 2 x$
$0 = 100 + 2 x$
$- 100 = 2 x$
$x = \frac{- 100}{2} = - 50$
Now, find the second derivative of the function $f (x)$ .
Use the second derivative test to determine whether the number we found was a critical number.
$f^{″} (x) = (100 + 2 x)^{'}$
$= (100)^{'} + (2 x^{'})$
$= 0 + 2$
$= 2 > 0$
Since, it is positive, it means that yes, there is a minimum.
Find y. Substitute $x = - 50$ into $y - x = 100$ .
$y + 50 = 100$
$y = 50$
Results: $x = - 50, y = 50$

Find two numbers whose difference is 100 and whose product is a minimum.

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