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Difference between revisions of "2007 iTest Problems/Problem 31"

(Created page with "== Problem == Let <math>x</math> be the length of one side of a triangle and let y be the height to that side. If <math>x+y=418</math>, find the maximum possible <math>\textit{i...")
 
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Let <math>x</math> be the length of one side of a triangle and let y be the height to that side. If <math>x+y=418</math>, find the maximum possible <math>\textit{integral value}</math> of the area of the triangle.  
 
Let <math>x</math> be the length of one side of a triangle and let y be the height to that side. If <math>x+y=418</math>, find the maximum possible <math>\textit{integral value}</math> of the area of the triangle.  
  
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== Solution ==
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==Solution==
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By an area formula for a triangle, the area of the triangle is <math>\frac{xy}{2}</math>.  Since <math>y = -x + 418</math>, substitute to get <math>\frac{-x^2 + 418x}{2} = -\frac{1}{2}x^2 + 209x</math>.
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The x-value to get the maximum is <math>\frac{-209}{2 \cdot -\frac{1}{2}} = 209</math>.  Thus, the maximum area of the triangle is <math>\frac{209^2}{2} = \frac{43681}{2} = 21840.5</math>, so the maximum integral area is <math>\boxed{21840}</math>.
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==See Also==
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{{iTest box|year=2007|num-b=30|num-a=32}}
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[[Category:Introductory Geometry Problems]]
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[[Category:Introductory Algebra Problems]]

Latest revision as of 18:27, 10 June 2018

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, Answer Key)
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Problem 30 		Followed by:
Problem 32
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