sparklemetink said:
I don't know how to set this up...
What is the largest possible area for a right triangle in which the sum of the lengths of the two shorter sides is 100 inches?
I know I need to get the perimeter first and then the area, but I don't know how to set this up except that a + b = 100.
Did you try drawing a diagram and labeling the two legs "a" and "b"? In a right triangle, the two legs are perpendicular (form a right angle)
The formula for the area of a triangle (generally) is this:
Area = (1/2)*(base)*(height)
In a right triangle, you can call one of the legs the base, and the other leg the height.
If you have identified the two legs as "a" and "b", then, for YOUR right triangle,
Area = (1/2)*a*b
Now...what else do you know? You know that the sum of the two shorter sides of this triangle is 100 inches. The two shorter sides are the legs, so you've correctly stated that
a + b = 100 inches
So, you could solve that for either a or b in terms of the other variable....I'll solve it for b in terms of a....
a + b = 100
a + b - a = 100 - a
b = 100 - a
We can now re-write the formula for the area, substituting (100 - a) for "b":
Area = (1/2)*a*b
Area = (1/2)*a*(100 - a)
or,
Area = 50 a - (1/2)a^2
I do not know what you have learned about finding maximum or minimum values.....
You've got the equation of a parabola here, one that opens downward. The maximum area occurs at the vertex of the parabola.
You'll need to give us some additional information about what techniques you have learned for dealing with this kind of problem.
