Difference between revisions of "2007 iTest Problems/Problem 31"
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==Solution== | ==Solution== | ||
| − | 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>. | + | 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>. | ||
==See Also== | ==See Also== | ||
Latest revision as of 18:27, 10 June 2018
Problem
Let 
be the length of one side of a triangle and let y be the height to that side. If 
, find the maximum possible 
of the area of the triangle.
Solution
By an area formula for a triangle, the area of the triangle is 
. Since 
, substitute to get 
.
The x-value to get the maximum is 
. Thus, the maximum area of the triangle is 
, so the maximum integral area is 
.
See Also
| 2007 iTest (Problems, Answer Key) | ||
| Preceded by: Problem 30 |
Followed by: Problem 32 | |
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