A wire of length 12 meters is bent into the shape of a rectangle. Find the dimensions of the rectangle that maximize its area.
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Let $l$ and $w$ be the length and width of the rectangle, respectively. Since the wire is bent into the shape of a rectangle, we know that $2l + 2w = 12$ or $l + w = 6$. We want to maximize the area $A$ of the rectangle, which is given by $A = lw$. To do this, we can solve for one variable in terms of the other and substitute it into the equation for area. Rearranging the equation $l + w = 6$, we find $l = 6 - w$. Substituting this into the equation for area, we have $A = (6 - w)w = 6w - w^2 = -(w^2 - 6w)$. Completing the square, we get $-(w^2 - 6w + 9) + 9 = -(w - 3)^2 + 9$. Since the square of a real number is at least 0, the maximum value of $(w - 3)^2$ is 0, and therefore the maximum value of $-(w - 3)^2$ is 0. The maximum value of $-(w - 3)^2 + 9 = 9$ and occurs when $w = \boxed{3}$. Then, $l = 6 - w = 6 - 3 = 3$. Thus, the dimensions of the rectangle that maximize its area are $\boxed{3}$ meters by $\boxed{3}$ meters.
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