Question

Find two numbers whose product is 100 and whose sum is a maximum?

Answer 1

$xy = 100$
$y = \frac{100}{x}$

$x + y = A, A \in \mathbb{R}$
$x + \frac{100}{x} = A$
$\frac{d}{dx}(x + \frac{100}{x}) = 0$, since the differential of a constant is just 0.

$1 - \frac{100}{x^{2}} = 0$, to find the stationary points.
$1 = \frac{100}{x^{2}}$
$x^{2} = 100$
$x = \pm 10$

$\frac{d^{2}}{dx^{2}}(x + \frac{100}{x}) = \frac{d}{dx}(1 - \frac{100}{x^{2}})$
$\frac{d}{dx}(1 - \frac{100}{x^{2}}) = \frac{200}{x^{3}}$

Thus, x = 10 will give you a maximum value, while x = -10 will give you a minimum value.

Thus, x and y = 10 will give you a maximum sum.

Answer 2

Let them be X and Y..
now...
yx=100 ....
y=100/y ...
now...
x+y= s....where s is their sum...
but y=100/x
now....x+100/x=s..
by differentiating both sides with respect to x..
1-100/x^2=ds/dx
but ds/dx=0..
100=x^2 ..
x=10..
so y=10