Asked by Anonymous
An isosceles triangle has a perimeter of 4 units. What should the length of its sides be so that the area is a maximum?
Answers
Answered by
oobleck
If it has a base of 2b and two sides of s, then
2b+2s = 4
The area
a = 1/2 bh = 1/2 b * √(s^2-b^2)
= 1/2 b √((2-b)^2 - b^2)
= 1/2 b √(4-4b)
= b√(1-b)
da/db = 0 when b=2/3
Note that this makes the triangle equilateral, as expected for maximum area. (just as a square has maximum area, and a cube has maximum volume)
2b+2s = 4
The area
a = 1/2 bh = 1/2 b * √(s^2-b^2)
= 1/2 b √((2-b)^2 - b^2)
= 1/2 b √(4-4b)
= b√(1-b)
da/db = 0 when b=2/3
Note that this makes the triangle equilateral, as expected for maximum area. (just as a square has maximum area, and a cube has maximum volume)
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