To solve this problem you must apply the proccedure shown below:
1- You have that the area rug must be no smaller that [tex] 24ft^{2} [/tex] and no bigger than [tex] 48ft^{2} [/tex] and its length is [tex] 2ft [/tex] more than its width.
2- The area of a rectangle can be calculated by multiplying the width by the length, if the length is [tex] 2ft [/tex] more than the width, you have that:
[tex] A=WL\\ L=W+2\\ A=W(W+2)=W^{2} +2W [/tex]
Where [tex] A [/tex] is the area, [tex] W [/tex] is the width and [tex] L [/tex] is the length.
3- Based on the information above, you can express it as below:
[tex] 24\leq W^{2} +2W\leq 48 [/tex]
4- You can separate to solve it if you want to see it better:
[tex] 24\leq W^{2}+2W\\ 0\leq W^{2}+2W-24\\ 4\leq W [/tex]
[tex] W^{2}+2W-48\leq 0\\ W\leq 6 [/tex]
5- Then, you have the following result:
[tex] 4\leq W\leq 6 [/tex]
6- And the length is:
[tex] 4+2\leq L\leq 6+2\\ 6\leq L\leq 8
[/tex]
Therefore, the answer is: [tex] 6\leq L\leq 8 [/tex]
The possible range for the value of length will be [tex]6\leq L \leq8[/tex]
According to the question we have:-
(1) The area rug should not be smaller [tex]24\ ft^2[/tex] than and not bigger [tex]45\ ft^2[/tex] than and its length is [tex]2\ ft[/tex] more than its width.
(2) The area of a rectangle is given as
[tex]A=L\times W[/tex]
Here [tex]L=W+2[/tex]
[tex]A=W(W+2)=W^2+2W[/tex]
Where A is the area, W is the width and L is the length.
(3) Based on the information above, you can express it as below:
[tex]24\leq W^2+2W \leq48[/tex]
(4) You can separate to solve it if you want to see it better:
[tex]24\leq W^2+2W[/tex]
[tex]0\leq W^2+2W-24[/tex]
[tex]4\leq W[/tex]
[tex]W^2+2W-48\leq 0[/tex]
[tex]W\leq 6[/tex]
(5) Then, you have the following result:
[tex]4\leq W\leq 6[/tex]
(6) Now the length will be calculated as
[tex]4+2\leq L\leq 6+2[/tex]
[tex]6\leq L\leq 8[/tex]
Thus possible range for the value of length will be [tex]6\leq L \leq8[/tex]
To know more about Area inequality follow
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