I am trying to pull out equations from word problems
P = 2L + 2W
P ≥ 692
L+ W ≥ 346
L+31≥ 346
L ≥315
This is simply an inequality problem.( ≥ ) should be used for at least ( minimum value),
The width of a rectangle is 31 centimeters. Find all possible values for the length of the rectangle if the perimeter is al least 692 centimerts
Tutors, please sign in to answer this question.
3 Answers
In a rectangle opposite sides are equal
perimeter = 2(length + Width)
2(l + W) must greater than or equal to 692
L + W must be greater than or equal to (692/2)346
L + 31 must be greater than or equal to 346
there fore L must be greater than or equal to (346 - 31) 315
:)
Good morning! Recall that perimeter of a rectangle is the sum of the lengths of all sides, so:
P= L + L + W + W, or P + 2L + 2W
If the width is 31 and total perimeter is at least 692, then the minimum would be:
692 = 2L + 2(31), or 692 = 2L + 62
Subtract 62 from both sides:
630 = 2L
Divide both sides by 2:
315 = L
Remember that the perimeter is at least 692; therefore the length would be at least 315 cm
Hope this helps
Mike