Solve the problem.
-One side of a rectangle is 2 times the other, and the perimeter is not to exceed 72. Find the possible values for x, the length of the shorter side.
A) $x \geq 24$
B) $x \leq 12$
C) $0

Question

Solve the problem.
-One side of a rectangle is 2 times the other, and the perimeter is not to exceed 72. Find the possible values for x, the length of the shorter side. 
A) $x \geq 24$
B) $x \leq 12$
C) $0 < x \leq 12$
D) $0 < x \leq 24$

Quizplus · Accepted Answer

The perimeter of a rectangle is given by $P = 2l + 2w$, where $l$ is the length and $w$ is the width. Given one side is twice the other, we can say $l = 2w$. Substituting $l$ with $2w$ in the perimeter formula gives $P = 2(2w) + 2w = 6w$. Since the perimeter must not exceed 72, we have $6w \leq 72$. Dividing both sides by 6 gives $w \leq 12$. Since $w$ represents the length of the shorter side and must be greater than 0, the possible values for $x$ are $0 < x \leq 12$.

Solve the Problem x≥24x \geq 24x≥24 B) x≤12x \leq 12x≤12 C) 0<x≤120 < x \leq 120<x≤12

Solve the Problem $x \geq 24$ B) $x \leq 12$ C) $0 < x \leq 12$