An isosceles triangle has a perimeter of 4 units. What should the length of its sides be so that the area is a maximum?
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)
To find the maximum area of an isosceles triangle with a given perimeter, we can apply the concept of calculus. Let's denote the length of the two congruent sides as "x" and the base as "y". Since it's an isosceles triangle, we have x = x, y = 2x, and the perimeter P = 2x + y.
Given that the perimeter is 4 units, we have:
2x + 2x = 4
4x = 4
x = 1
So, the length of the two congruent sides is 1 unit, and the base is 2 units.
To find the maximum area, we can use the formula for the area of an isosceles triangle: Area = (1/2) * y * h, where "h" is the height of the triangle.
In this case, "h" is also the perpendicular bisector of the base, dividing it into two equal parts. Therefore, the length of "h" can be found using the Pythagorean theorem:
h^2 = x^2 - (y/2)^2
h^2 = 1^2 - (2/2)^2
h^2 = 1 - 1/4
h^2 = 3/4
h = sqrt(3/4)
h = sqrt(3) / 2
Now we can substitute the values into the formula for the area:
Area = (1/2) * y * h
Area = (1/2) * 2 * (sqrt(3) / 2)
Area = sqrt(3)/2
Hence, the area of the isosceles triangle with the maximum area, given the perimeter of 4 units, is sqrt(3)/2 square units.
To find the length of the sides of an isosceles triangle with a maximum area, we need to use calculus. We'll start by defining the variables.
Let's assume the length of the equal sides of the triangle is 'a', and the length of the base is 'b'. Since the triangle is isosceles, the two equal sides are of length 'a', and the base is of length 'b'.
Given that the perimeter of the triangle is 4 units, we have the equation:
2a + b = 4
Now, we need to express the area of the triangle (A) in terms of 'a' and 'b'. The formula for the area of a triangle is:
A = (1/2) * base * height
In an isosceles triangle, the height is perpendicular to the base and bisects it, forming two right-angled triangles. Let's call the height 'h'.
Since the equal sides are 'a', we now have two identical right-angled triangles. So, in one triangle, the base is 'b/2', and the height is 'h'. Using the Pythagorean theorem, we have:
h^2 + (b/2)^2 = a^2
Rearranging the equation, we can express the height in terms of 'a' and 'b' as:
h = sqrt(a^2 - (b/2)^2)
Now, we can express the area of the triangle in terms of 'a' and 'b' as:
A = (1/2) * b * sqrt(a^2 - (b/2)^2)
To find the maximum value of 'A', we need to find the critical points by taking the partial derivatives of 'A' with respect to 'a' and 'b'. We'll differentiate 'A' with respect to 'b' first:
dA/db = (1/2) * (sqrt(a^2 - (b/2)^2) - (b/2) * (2b / 2 * sqrt(a^2 - (b/2)^2))
Simplifying this expression, we get:
dA/db = (1/2) * (sqrt(a^2 - (b/2)^2) - (b^2 / sqrt(a^2 - (b/2)^2))
To find the maximum value, we set the derivative equal to zero:
(1/2) * (sqrt(a^2 - (b/2)^2) - (b^2 / sqrt(a^2 - (b/2)^2) = 0
Squaring both sides of the equation, we get:
3(a^2 - (b/2)^2) = b^2
Expanding this equation, we get:
3a^2 - 3(b/2)^2 = b^2
Simplifying, we have:
3a^2 - (9/4)b^2 = 0
Now, let's differentiate 'A' with respect to 'a':
dA/da = (1/2) * (b / sqrt(a^2 - (b/2)^2)
Again, setting the derivative equal to zero, we have:
b / sqrt(a^2 - (b/2)^2) = 0
This equation implies that b must be zero, which is not possible since it represents the length of one of the sides of the triangle.
Therefore, we only need to solve the equation 3a^2 - (9/4)b^2 = 0 to find the length of the sides 'a' and 'b' that maximize the area.
Solving this equation, we get:
a = b * sqrt(3/9) = b * sqrt(1/3)
Now, using the perimeter equation 2a + b = 4, we can substitute the value of 'a' to find 'b' and then find 'a'.
2(sqrt(1/3))b + b = 4
Simplifying, we get:
2(sqrt(1/3) + 1)b = 4
(sqrt(1/3) + 1)b = 2
b = 2 / (sqrt(1/3) + 1)
Substituting the value of 'b' back into the equation for 'a', we have:
a = (2 / (sqrt(1/3) + 1)) * sqrt(1/3)
Now, we have the values of 'a' and 'b' that maximize the area of the isosceles triangle.