The diagram shows a line with intercepts 3 and 9 and one of many rectangles which can be inscribed under this line and within the first quadrant. a. write an equation for the line. b. determine if the rectangle can have x = 4 and y = 2 as its dimensions. c. find x and y if the rectangle is a square. d. write an expression for a, the area of the rectangle, using x as the only variable. e. find x if the area of the rectangle is 6. f. which x gives the rectangle its largest area?

A) The easiest way to write an equation for this line is to use intercept form:
.. x/x-intercept +y/y-intercept = 1
.. x/9 +y/3 = 1

b) The point (x, y) = (4, 2) does not satisfy the equation. (see the graph) These cannot be the dimensions of an inscribed rectangle.

c) The intersection with the line y=x is (x, y) = (2.25, 2.25). These are the dimensions of an inscribed square.

d) The equation can be rewritten as
.. x +3y = 9
.. y = 3 -(x/3)
Then the area can be written as
.. A = xy = x(3 -x/3)

e) For x=6, the area is
.. A = 6(3 -6/3) = 6*1 = 6 . . . . square units

f) The parabola described by A has zeros at x=0 and x=9. The vertex of that parabola is on the line of symmetry, at x = 4.5. That value of x gives the maximum area.

The answer is $9.90 with rounding to the nearest tenth