A rectangle is to be constructed so that it is bounded by the x-axis, the y-axis, and the line y = -3x + 4. What is the maximum area of this rectangle?

the y-intercept is 4, so the height is 4

the x-intercept is 4/3, so the base is 4/3

area = (1/2)base x height
= (1/2)(4)(4/3) = 8/3 square units

To find the maximum area of the rectangle, we need to understand the geometric properties of the rectangle and how it is bounded by the given constraints.

Let's start by visualizing the problem. We have the x-axis (y = 0), the y-axis (x = 0), and the line y = -3x + 4. The rectangle will be bounded by these three.

To find the maximum area, we need to find the dimensions of the rectangle that will maximize its area. Let's assume the length of the rectangle is along the x-axis and has a width along the y-axis.

To construct this rectangle, we need to find the x-coordinate where the line y = -3x + 4 intersects with the x-axis. We can set y equal to zero and solve for x:

0 = -3x + 4
3x = 4
x = 4/3

So, the x-coordinate where the line intersects with the x-axis is (4/3, 0).

Similarly, to find the y-coordinate where the line y = -3x + 4 intersects with the y-axis, we can set x equal to zero:

y = -3(0) + 4
y = 4

So, the y-coordinate where the line intersects with the y-axis is (0, 4).

Now we have two opposite vertices of the rectangle: (4/3, 0) and (0, 4). To maximize the area, the rectangle should be a square. In a square, all sides are equal.

The length of the rectangle is the difference in x-coordinates: 4/3 - 0 = 4/3.
The width of the rectangle is the difference in y-coordinates: 4 - 0 = 4.

Since the rectangle has equal sides, the side length of the square is the square root of the product of the length and width:

Side length = sqrt(4/3 * 4) = sqrt(16/3) = 4/sqrt(3) = (4/√3).

Therefore, the maximum area of the rectangle is obtained by squaring the side length:

Area = (4/√3) * (4/√3) = 16/3.

So, the maximum area of the rectangle bounded by the x-axis, the y-axis, and the line y = -3x + 4 is 16/3.