A rectangle has a length of (8-2x) and a width of 5. Find x so that the area of the rectangle is less than 30 by writing and solving a compound inequality.

Question

Compound inequality?

Answer 1

A compound inequality has at least 2 inequalities that are separated by "and" or "or".  For example, if you wanted to write that the absolute value of x was less than 6, you could write "x > -6 and x < 6" or you could write "-6 < x < 6".  These are equivalent forms of the same compound inequality.  If you wanted to write that the absolute value of x was greater than 6, you would write "x < -6 or x > 6".  

 

Now on to the rectangle problem.  The area of rectangle is length times width, so we have

 

A = (8-2x)*5 = 40 - 10x

 

We want the area to be less than 30, and we also know that the area has to be greater than 0 (this is where the compound part comes in).  So we can write 

 

0 < A < 30

0 < 40 -10x < 30

-40 < -10x < -10

4 > x > 1               Remember to switch the sense of the inequality when you divide through by a negative number

 

This is usually written from low to high as 1 < x < 4

 

You can check your answer by plugging in the endpoints.  When x = 1 then 8-2x = 6 and the rectangle is 6 by 5 with an area of 30.  Making x bigger than 1 will make 8-2x smaller so the area will be less than 30.  You can keep increasing x all the way up to 4 at which point the length becomes 8-2(4) = 0 and the rectangle disappears.