[tex]xy = 100[/tex]
[tex]y = \frac{100}{x}[/tex]
[tex]x + y = A, A \in \mathbb{R}[/tex]
[tex]x + \frac{100}{x} = A[/tex]
[tex]\frac{d}{dx}(x + \frac{100}{x}) = 0[/tex], since the differential of a constant is just 0.
[tex]1 - \frac{100}{x^{2}} = 0[/tex], to find the stationary points.
[tex]1 = \frac{100}{x^{2}}[/tex]
[tex]x^{2} = 100[/tex]
[tex]x = \pm 10[/tex]
[tex]\frac{d^{2}}{dx^{2}}(x + \frac{100}{x}) = \frac{d}{dx}(1 - \frac{100}{x^{2}})[/tex]
[tex]\frac{d}{dx}(1 - \frac{100}{x^{2}}) = \frac{200}{x^{3}}[/tex]
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.
