Q:

# The perimeter of a rectangle is at most 80 inches. The length of the rectangle is 25 inches. The inequality 80-2w (is greater than or equal to) 50 , can be used to find w, the width of the rectangle in inches. Solve the inequality and interpret the solution. How will the solution change if the width must be at least 10 inches and a whole number? I know how to solve the equation, it is the second part I'm stumped on.

Accepted Solution

A:
Let's set up a system of equations to find the inequality, and then move on to the second part.
We know the perimeter of a rectangle P = 2w + 2l, where w is width and l is length. Therefore:
2w + 2l ≤ 80
We are given l, which makes our job easier
2w + 2(25) ≤ 80
2w ≤ 30
w ≤ 15.

If the width must be at least 10 inches and whole number, the inequality would change in this way:
10 ≤ w ≤ 15 , with
{10, 11, 12, 13, 14, 15}
the set of possible values, or the domain, of w.

In terms of the perimeter P, we would know that the perimeter is subject to:
2(10) + 2(25) ≤ P ≤ 2 (15) + 2(25)
70 ≤ P ≤ 80
When we consider our constraint for w (must be a whole number), P can actually only have 6 values:
P: {70, 72, 74, 76, 78, 80}

If we map our two variables W and P, with W the independent variable (x-coordinate) and P the dependent variable (y-coordinate), our solution set is:
(W,P) = {(10,70), (11,72), (12, 74), (13,76), (14,78), (15,80)}