The perimeter of a rectangle is 40 feet. Describe the possible lengths of a side if the area of the rectangle is not to exceed 84 square feet.
Interval notation?
- Posted:
- 3+ months ago by draco64114
- Topics:
- square, side, feet, area, perimeter, mathematics, length
Details:
Answers (2)
This is a simulltaneous system. You have two equations, one for perimeter and one for area.
p = 40 = 2x + 2y
a = 84 = xy
You have two lines on a plot. Where they cross, the values of x and y satisfy both equations simultaneously, so it is called a simultaneous system. Like this:
www.wolframalpha.com/input/?i=solve+40+%3D+2x+%2B+2y%2C+84+%3D+xy
There are two ways to find that point. Substitution is when you solve one equation for one variable and substitute that into the other equation.
84 = xy
y = 84/x Substitute this for y in the other equation.
40 = 2x + 2(84/x) Divide by 2.
20 = x + 84/x Multiply by x. Any time you multiply or divide by the variable, you have to make an extra step to be sure all your results are valid in the original equations.
20x = x^2 + 84 Subtract 20x.
x^2 - 20x + 84 = 0 Factor,
(x - 6)(x - 14) = 0
x = 6, 14
Elimination is when you add the equations in a way to eliminate one variable. Otherwise the process is the same. (It doesn't work in this example.)
p = perimeter
a = area
Let's begin with a square. The perimeter would then be:
p = 4 * 10 = 40, or
p = 2 (10 + 10) = 40. I know, this looks strange, but bear with me.
The perimeter of a rectangle other than a square can be described as:
p = 2 (10 - x + 10 + x) with the two orthogonal sides 10-x and 10+x.
The area of this rectangle is:
a = (10 - x) (10 + x)
a = 100 - x²
For x >= 10 you won't get a rectangle.
If x = 0, then a = 10 * 10 = 100, which is far too large,
if the area shall not exceed 84 ft². So, for a = 84
84 = 100 - x²
x² = 16
x = 4. We are only interested in positive lengths.
For x < 4 the rectangle would be too large.
So, your set of solutions is
S = {x | 4 <= x < 10} = [4, 10) .