A triangle is drawn inside a bounded region of y² = 2x and x = 24. Then maximum area of triangle is

Question

\(ABCD\) is a parallelogram such that \(AF = 7 \text{ cm}\), \(FB = 3 \text{ cm}\) and \(EF = 4 \text{ cm}\), length \(FD\) equals

Answer 1

To find the maximum area of a triangle drawn inside the region bounded by the parabola y^2 = 2x and the line x = 24, let's proceed step-by-step.

The parabola y^2 = 2x opens to the right, and the line x = 24 is a vertical line. The feasible region is where the parabola intersects this line, which means the maximum x-coordinate of the triangle is 24, i.e., x = 24.
Substitute x = 24 into the parabola equation to find the corresponding y values:
y^2 = 2 \times 24 = 48 \quad \Rightarrow \quad y = \pm \sqrt{48} = \pm 4\sqrt{3}
This means the parabola intersection points on this vertical line are (24, 4\sqrt{3}) and (24, -4\sqrt{3}).
The base of the triangle will be along this line x = 24 from y = -4\sqrt{3} to y = 4\sqrt{3}, giving a base length of 8\sqrt{3}.
The vertex of the triangle, the maximum possible, should be at the vertex of the parabola, which occurs at the focus (maximum horizontal displacement at the minimum y-coordinate), meaning the maximum height when x = 0. For y = 0, the vertex (or focus point) of this parabola is (0, 0).
Using the formula for the area of a triangle with base B and height h:
A = \frac{1}{2} \times B \times h
Substituting the maximum base 8\sqrt{3} and maximum height 24:
A = \frac{1}{2} \times 8\sqrt{3} \times 24 = 96\sqrt{3}

Therefore, the maximum area of the triangle that can be inscribed in the given region is 96√3.

Answer 2

To find the maximum area of a triangle drawn inside the region bounded by the parabola y^2 = 2x and the line x = 24, let's proceed step-by-step.

The parabola y^2 = 2x opens to the right, and the line x = 24 is a vertical line. The feasible region is where the parabola intersects this line, which means the maximum x-coordinate of the triangle is 24, i.e., x = 24.
Substitute x = 24 into the parabola equation to find the corresponding y values:
y^2 = 2 \times 24 = 48 \quad \Rightarrow \quad y = \pm \sqrt{48} = \pm 4\sqrt{3}
This means the parabola intersection points on this vertical line are (24, 4\sqrt{3}) and (24, -4\sqrt{3}).
The base of the triangle will be along this line x = 24 from y = -4\sqrt{3} to y = 4\sqrt{3}, giving a base length of 8\sqrt{3}.
The vertex of the triangle, the maximum possible, should be at the vertex of the parabola, which occurs at the focus (maximum horizontal displacement at the minimum y-coordinate), meaning the maximum height when x = 0. For y = 0, the vertex (or focus point) of this parabola is (0, 0).
Using the formula for the area of a triangle with base B and height h:
A = \frac{1}{2} \times B \times h
Substituting the maximum base 8\sqrt{3} and maximum height 24:
A = \frac{1}{2} \times 8\sqrt{3} \times 24 = 96\sqrt{3}

Therefore, the maximum area of the triangle that can be inscribed in the given region is 96√3.

The Correct Option is C