Answer:
width must be atleast 23 inches.
Step-by-step explanation:
Given that the length of a rectangle is 6 inches longer than its width.
If l is length, then w = l-6
Area =lw
i.e. Area [tex]= l(l-6)\geq 667\\l^2-6l-667\geq 0\\(l-29)(l+23)\geq 0[/tex]
For this inequality being the product of two numbers is positive if both terms have the same sign.
This is possible only if l is atleast 29 inches (ignoring negative solution for l)
Hence possible widths are atleast 23 inches.
