Question

An isosceles triangle has its vertex at the origin and its base parallel to the x- axis with the vertices above the axis on the curve y = 27-x². Find the largest area the triangle can have.

Answer 1

The largest area of the isosceles triangle is 243/2 square units.

To find the largest area of the isosceles triangle with its vertex at the origin and base parallel to the x-axis, we consider the vertices of the triangle lying above the x-axis on the curve y=27−x^2. The base of the triangle is formed by two points where y=0, giving us 27−x^2=0. Solving for x, we find x=±3.

The base length of the triangle is 2×3=6 units. The height of the triangle, which is the y-coordinate of the curve at the origin, is 27. The area of the triangle is given by 1/2×base×height, and substituting the values, we get 1/2×6×27= 243/2 square units. Therefore, the largest possible area of the isosceles triangle is 243/2 square units.