Question

The length of a rectangle is 15 and its width is w.

The perimeter of the rectangle is, at most, 50,
Which inequality can be used to find the longes
possible width?
A
30+ 2x < 50
B. 30 + 2w < 50
D. 30+ 2w > 50
C.30 + 2w > 50

Answer 1

2x+30< 50 ...answer : A

Explanation:

the perimeter is : P = 2(L+W)

let : L= x given : L=15

P= 2(x+15)

The perimeter of the rectangle is, at most, 50: P<50

2(x+15) < 50

2x+30< 50 ...answer : A

Answer 2

A rectangle with sides 15 and w has perimeter

`2p = 2(15+w) = 2w+30`

We want this quantity to be at most 50, so it must be less than or equal to 50:

`2w+30\leq 50`

For the record, this implies that

`2w\leq 20 \iff w \leq 10`

So, the width can be at most 10.