Find the volume of the pyramid bounded by the plane x + 2y + 6z = 12 in the ﬁrst octant(x ≥ 0, y ≥ 0, z ≥ 0)

Question

Find the volume of the pyramid bounded by the plane $x + 2y + 6z = 12$ in the first octant ( $x \geq 0, y \geq 0, z \geq 0$ )

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Solution

To find the volume of the pyramid, we first need to find the intercepts of the plane with the x, y, and z axes.

The x-intercept is found by setting y = 0 and z = 0 in the equation of the plane, which gives us x = 12.

The y-intercept is found by setting x = 0 and z = 0 in the equation of the plane, which gives us y = 6.

The z-intercept is found by setting x = 0 and y = 0 in the equation of the plane, which gives us z = 2.

The volume V of a pyramid is given by the formula V = 1/3 * base area * height. In this case, the base of the pyramid is a triangle with vertices at the origin and the x and y-intercepts, and the height is the z-intercept.

The area A of the base triangle is given by the formula A = 1/2 * base * height. In this case, the base is the distance from the origin to the x-intercept (12 units), and the height is the distance from the origin to the y-intercept (6 units), so A = 1/2 * 12 * 6 = 36 square units.

Therefore, the volume of the pyramid is V = 1/3 * 36 * 2 = 24 cubic units.

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Find the volume of the pyramid bounded by the plane x + 2y + 6z = 12 in the ﬁrst octant(x ≥ 0, y ≥ 0, z ≥ 0)

Question

Find the volume of the pyramid bounded by the plane x+2y+6z=12 x + 2y + 6z = 12 x+2y+6z=12 in the first octant (x≥0,y≥0,z≥0 x \geq 0, y \geq 0, z \geq 0 x≥0,y≥0,z≥0)

Solution

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Find the volume of the pyramid bounded by the plane $x + 2y + 6z = 12$ in the first octant ( $x \geq 0, y \geq 0, z \geq 0$ )