2007 iTest Problems/Problem 31

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Problem

Let $x$ be the length of one side of a triangle and let y be the height to that side. If $x+y=418$, find the maximum possible $\textit{integral value}$ of the area of the triangle.

Solution

By an area formula for a triangle, the area of the triangle is $\frac{xy}{2}$. Since $y = -x + 418$, substitute to get $\frac{-x^2 + 418x}{2} = -\frac{1}{2}x^2 + 209x$.

The x-value to get the maximum is $\frac{-209}{2 \cdot -\frac{1}{2}} = 209$. Thus, the maximum area of the triangle is $\frac{209^2}{2} = \frac{43681}{2} = 21840.5$, so the maximum integral area is $\boxed{21840}$.

See Also

2007 iTest (Problems)
Preceded by:
Problem 30
Followed by:
Problem 32
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