The length of a rectangle is three feet less than twice its width. If x represents the width of the rectangle, in feet, which inequality represents the area of the rectangle that is at most 30 square feet?

Options

A. `x(2x-3)le30`
B. `x(2x-3)ge30`
C. `x(3-2x)le30`
D. `x(3-2x)ge30'

Answer

To solve the problem step by step, we need to define the variables and set up the inequality based on the information given.

### Step 1: Define the Variables
Let \( x \) represent the width of the rectangle in feet.

### Step 2: Express the Length in Terms of Width
According to the problem, the length of the rectangle is three feet less than twice its width. Therefore, we can express the length \( L \) as:
\[ L = 2x - 3 \]

### Step 3: Write the Expression for the Area
The area \( A \) of a rectangle is given by the formula:
\[ A = \text{length} \times \text{width} \]
Substituting the expressions we have:
\[ A = x \cdot (2x - 3) \]

### Step 4: Set Up the Inequality for the Area
We are told that the area of the rectangle is at most 30 square feet. This can be expressed as:
\[ x(2x - 3) \leq 30 \]

### Step 5: Rearrange the Inequality
To make it clearer, we can rearrange the inequality:
\[ 2x^2 - 3x \leq 30 \]
Subtract 30 from both sides:
\[ 2x^2 - 3x - 30 \leq 0 \]

### Step 6: Identify the Correct Inequality
The inequality that represents the area of the rectangle that is at most 30 square feet is:
\[ 2x^2 - 3x - 30 \leq 0 \]

### Final Answer
Thus, the correct inequality representing the area of the rectangle that is at most 30 square feet is:
\[ x(2x - 3) \leq 30 \]