What is the maximum area of a rectangle that has sides along the positive x-axis and y-axis and lies below the line 3x+2y=1

Let the right-hand corner vertex be P(x,y)

but 3x+2y = 1
y = (1-3x)/2 or 1/2 - (3/2)x

area = xy
= x(1/2 - (3/2)x)
= (1/2)x - (3/2)x^2

d(area)/dx = 1/2 - 3x
= 0 for a max of area
3x = 1/2
x = 1/6
y = 1/2 - (3/2)(1/6) = 1/4 ---> P is (1/6 , 1/4)

max area = xy = (1/6)(1/4) = 1/24 square units

To find the maximum area of a rectangle with sides along the positive x-axis and y-axis, we need to consider the geometric constraints imposed by the given line.

Let's break down the problem into steps:

1. Determine the vertices of the rectangle:
We know that one side of the rectangle lies along the x-axis, so its length will be equal to the x-coordinate of the vertex opposite to it. Similarly, the other side of the rectangle lies along the y-axis, so its length will be the y-coordinate of the vertex opposite to it.

Let's find the coordinates of the two vertices where the line 3x + 2y = 1 intersects the x-axis and y-axis, forming the rectangle:

When the line intersects the x-axis, y is zero:
3x + 2(0) = 1
3x = 1
x = 1/3

When the line intersects the y-axis, x is zero:
3(0) + 2y = 1
2y = 1
y = 1/2

So the two vertices of the rectangle are (1/3, 0) and (0, 1/2).

2. Calculate the area of the rectangle:
The area of a rectangle is given by the product of its length and width. In this case, the length of the rectangle is the x-coordinate of the vertex on the y-axis, and the width is the y-coordinate of the vertex on the x-axis.

Length = x-coordinate of (0, 1/2) = 0
Width = y-coordinate of (1/3, 0) = 0

Therefore, the area of the rectangle is zero.

In conclusion, the maximum area of the rectangle that lies below the line 3x + 2y = 1 is zero, since the rectangle formed by the intersection points on the x-axis and y-axis has zero area.