2007 iTest Problems/Problem 31

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 $\boxed{21840}$ .

2007 iTest (Problems, Answer Key)
Preceded by: Problem 30	Followed by: Problem 32
1 • 2 • 3 • 4 • 5 • 6 • 7 • 8 • 9 • 10 • 11 • 12 • 13 • 14 • 15 • 16 • 17 • 18 • 19 • 20 • 21 • 22 • 23 • 24 • 25 • 26 • 27 • 28 • 29 • 30 • 31 • 32 • 33 • 34 • 35 • 36 • 37 • 38 • 39 • 40 • 41 • 42 • 43 • 44 • 45 • 46 • 47 • 48 • 49 • 50 • 51 • 52 • 53 • 54 • 55 • 56 • 57 • 58 • 59 • 60 • TB1 • TB2 • TB3 • TB4

Page

Toolbox

Search

2007 iTest Problems/Problem 31

Problem

Solution

See Also