Competition problem — Algebra — Maths Olympiad Prep

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Algebra Difficulty 5.4 AIME, harder Prove it Philippines

Problem:

$x + \frac{1}{x}$ has a maximum in $x < 0$ and a minimum in $x > 0$ . Find the area of the rectangle whose sides are parallel to the axes and two of whose vertices are the maximum and minimum values of $x + \frac{1}{x}$ .

Solution

Solution:

(ans. area $= 4 \cdot 2 = 8$ .
Vertex at maximum is $(-1, -2)$ , vertex at minimum is $(1, 2)$ . Thus, width is $2$ and height is $4$ . Vertices are obtained from the inequalities $\frac{(x+1)^2}{x} \geq 0$ for $x > 0$ and $\frac{(x-1)^2}{x} \leq 0$ for $x < 0$ .)

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