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EvoBruh3273

12.05.2023

Mathematics

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

$x + y = A, A \in \mathbb{R}$
$x + \frac{100}{x} = A$
$\frac{d}{dx}(x + \frac{100}{x}) = 0$ , 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.

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