Draw a rectangle that is twice as wide as it is tall, and that fits snugly into the triangular region formed by the line 3x + 4y = 12 and the positive coordinate axes, with one corner at the origin and the opposite corner on the line. find the dimensions of this rectangle.

To draw the line 3x+4y=12:

let x=0, then      4y=12, so y=3                 thus the point (0, 3) is on the line

let y-0, then       3x=12, so x=4                 thus the point (4, 0) is on the line

Join the points to form the line as shown in the picture.

Let the opposite corner of the origin be P, the coordinates of P must be (2a, a). 

In this way we have "a rectangle that is twice as wide as it is tall"

the point P(2a, a) is a point on the line 3x+4y=12, 

thus x=2a, y=a is a solution to the equation:

3x+4y=12
3(2a)+4a=12
6a+4a=12
10a=12
a=12/10=1.2

The dimensions of the rectangle are 2a by a, that is 2.4 by 1.2

2.4 by 1.2

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