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3 votes
The length of a rectangle is 6 inches longer than its width. what are the possible widths if the area of the rectangle is at least 667 square inches

2 Answers

4 votes
area=l×w
l=w+6
667=w(w+6)
w²+6w-667=0
w=23 or -29
w=23 only because distance cannot be negative
User Imperalix
by
8.6k points
1 vote

Answer:

width must be atleast 23 inches.

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
= l(l-6)\geq 667\\l^2-6l-667\geq 0\\(l-29)(l+23)\geq 0

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.

User MCM
by
8.1k points
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