The length of a rectangle is 5 times the width. If the perimeter is to be less than 84 meters. What are possible values for width?

Question

I didn't realize it was going to ask you to set it in interval notation, which is very confusing.

Answer 1

L = 5w

84 > Perimeter = 2L + 2w = 2(5w) + 2w = 10w+2w = 12w

7 > w

w < 7

(0, 7) <--- width cannot be negative, nor zero

Answer 2

Given a rectangle with a legnth (L) equal to five times the width (W), and the perimeter is less than 84 m.

Find the set of all possible values for the width (W) and express the set using interval notation.

Equation-1 L = 5*W (Given)

Equation-2 P = 2*L + 2*W (Definition of rectangle perimeter)

Inequality-1 P < 84 (Given)

Step 1 Substitute Equation-1 into Equation-2 (replacing L with 5*W):

Eq#2: P = 2 * L + 2*W

rewrite: P = 2 * (5*W) + 2*W

Step 2 Simplify the rewritten Equation.

P = 2* (5*W) + 2*W (associative law)

P = (2*5)*W + 2*W (using 2*5=10)

P = 10*W + 2*W (using distributive law)

* P = 12*W

Step 3 Substitute Result (*) of Step 2 (P=12*W) into Inequality-1 (P < 84)--replacing P with 12*W.

Ineq#1: P < 84 meters

rewrite: 12*W < 84 meters

Step 4 Simplify the rewritten Inequality

12*W < 84 (using 84÷12=7)

W < 7

-------------------DONE---------------------------

W < 7 limits the values of W because a rectangle may not have zero-length sides.

Therefore, the solution is 0 < W < 7 . . . which must be written using interval notation.

This is an open-ended interval--where W cannot be equal to the endpoints, i.e. the endpoints are

not in the set of possible values of W...

This type of interval is represented with open parentheses (rather than closed brackets)

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Inequality 0 < W < 7

Interval Notation: ( 0 , 7 )

Set Builder Notation: { W | 0 < W < 7 }

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Jamie Bernhard

(www.wyzant.com/tutors/JmeB)

Answer 3

We are given that the length L is 5 times the width W, so L=5W. The area is less than 84m and is calculated by L*W, so 84>L*W=5W*W=5W² so 5W²<84 -> W²<16.8 -> W<4.09878 so 0<W<4.09878