To find the area of the part of the plane described by the equation 5x + 4y + z = 20 that lies in the first octant, we need to determine the boundaries of this plane in relation to the axes.
First, let’s rewrite the equation to express z in terms of x and y:
z = 20 – 5x – 4y
In the first octant, all coordinates must be non-negative (i.e., x ≥ 0, y ≥ 0, z ≥ 0). This gives us constraints:
- When y = 0: 5x + z = 20 8> z = 20 – 5x 8> The x-intercept occurs when x = 0, giving z = 20, and x = 4 gives z = 0.
- When x = 0: 4y + z = 20 8> z = 20 – 4y 8> The y-intercept occurs when y = 0, giving z = 20, and y = 5 gives z = 0.
Next, to find the area bounded by these intercepts on the xy-plane (which can be visualized as a triangle), we find the vertices of the triangle formed by the points (0, 0, 20), (4, 0, 0), and (0, 5, 0).
To calculate the area of the triangle in the xy-plane, we need the base and height of the triangle:
- Base = 4 (from (0, 0, 0) to (4, 0, 0))
- Height = 5 (from (0, 0, 0) to (0, 5, 0))
The area of a triangle is calculated using the formula:
Area = 0.5 × base × height
Area = 0.5 × 4 × 5 = 10
Thus, the area of the part of the plane 5x + 4y + z = 20 that lies in the first octant is 10 square units.