An isosceles triangle is drawn with its vertex at the origin and its base parallel to the x-axis. The vertices of the base are on the curve 5y=25-x^2 Find the area of the largest such triangle.

your parabola is y = 5 - (1/5)x^2

let the points of contact of the base of the triangle be (x,y) and (-x,y)
Then the area is
Area = xy/2
= (x/2)(5-(1/5)x^2)
= 5x/2 - (1/10)x^3

d(Area)/dx = 5/2 - 3x^2/10
= 0 for a max of Area
3x^2/10 = 5/2
x^2 = 25/3
x = ± 5/√3 or appr. 2.88675

sub back in area expression
I had appr. 4.81125

( I tried x = 3 and x = 2.8 and they both gave me slightly smaller areas)

The base of the triangle should have been 2x, which makes the

Area = xy
= 5x - (1/5)x^3

Can you figure out how that changes the answers?
(hint: one stays the same, ...)

goodexcelent enteligent proud on u

To find the area of the largest isosceles triangle, we need to determine the length of the base and the height of the triangle.

Given that the base is parallel to the x-axis, the equation of the curve can be rearranged to solve for y in terms of x:

5y = 25 - x^2
y = (25 - x^2) / 5

Since the triangle is isosceles, the distance from the origin (the vertex) to the x-coordinate of the base will be the same as the distance from the origin to the x-coordinate of the top vertex. Let's call this x value "a."

Using this information, we can find the length of the base of the triangle by subtracting the x-coordinate of the base's right vertex from the x-coordinate of the base's left vertex:

base = a - (-a) = 2a

To find the height of the triangle, we need to determine the y-coordinate of the top vertex. Since the top vertex lies on the curve, we substitute x = a into the equation for y:

y = (25 - a^2) / 5

The height of the triangle is the distance between the y-coordinate of the top vertex and the y-coordinate of the origin. So the height is:

height = (25 - a^2) / 5 - 0 = (25 - a^2) / 5

Now we can find the area of the triangle by using the formula for the area of a triangle:

Area = (base * height) / 2

Substituting the values we found:

Area = (2a * (25 - a^2) / 5) / 2

Simplifying further:

Area = a * (25 - a^2) / 5

To find the maximum area, we can differentiate the area equation with respect to a and set it equal to zero:

d(Area)/da = (25 - 3a^2) / 5 = 0

Solving for a:

25 - 3a^2 = 0
3a^2 = 25
a^2 = 25/3
a = sqrt(25/3) or -sqrt(25/3)

Since the length of a side cannot be negative, we consider the positive value of a:

a = sqrt(25/3)

Now we substitute this value of a into the area equation to find the maximum area:

Area = (sqrt(25/3) * (25 - (sqrt(25/3))^2)) / 5

Simplifying further:

Area = (sqrt(25/3) * (25 - 25/3)) / 5
Area = (sqrt(25/3) * (75/3 - 25/3)) / 5
Area = (sqrt(25/3) * 50/3) / 5
Area = (50/3)^(1/2) * 50/3 * 1/5
Area = 10/3 * (50/3) * 1/5
Area = 500/45
Area ≈ 11.111 square units

Therefore, the area of the largest isosceles triangle is approximately 11.111 square units.