Unit Pretest Select the correct answer. A carpenter is building a rectangular table. He wants the perimeter of the tabletop to be no more than 28 feet. He also wants the lel be greater than or equal to the square of 2 feet less than its width. Create a system of inequalities to model the situation, where $$ x $$ represents the width of the tabletop and $$ y $$ represents the length of tl this system of inequalities to determine the viable solutions. A. Part of the solution region includes a negative length; therefore, not all solutions are viable for the given situation. B. The entire solution region is viable. C. Part of the solution region includes a negative width; therefore, not all solutions are viable for the given situation. D. No part of the solution reaion is viable because the lenath and width cannot be neaative.

Question

UpStudy AI · Accepted Answer

To model the carpenter's situation, let's define the variables and set up the inequalities based on the given conditions:

- **Variables:**
  - $$ x $$: Width of the tabletop (in feet)
  - $$ y $$: Length of the tabletop (in feet)

- **Conditions:**
  1. **Perimeter Constraint:** The perimeter of a rectangle is given by $$ 2x + 2y $$. The carpenter wants the perimeter to be no more than 28 feet:
     $$
     2x + 2y \leq 28
     $$
     Simplifying:
     $$
     x + y \leq 14
     $$
  
  2. **Length Constraint:** The carpenter wants the length $$ y $$ to be at least the square of 2 feet less than its width:
     $$
     y \geq (x - 2)^2
     $$

- **System of Inequalities:**
  $$
  \begin{cases}
    x + y \leq 14 \
    y \geq (x - 2)^2
  \end{cases}
  $$

- **Analyzing the Solution Region:**
  - **Graphical Interpretation:**
    - The inequality $$ x + y \leq 14 $$ represents a region below the line $$ y = 14 - x $$.
    - The inequality $$ y \geq (x - 2)^2 $$ represents a region above the parabola $$ y = (x - 2)^2 $$.

- **Intersection Points:** Solving $$ (x - 2)^2 = 14 - x $$ gives $$ x = 5 $$ and $$ x = -2 $$. However, $$ x = -2 $$ (a negative width) is not physically meaningful for the tabletop.

- **Viable Solutions:** While there are points that satisfy both inequalities, some of these points involve negative values for $$ x $$, which would imply a negative width. Since negative dimensions are not feasible in this context, not all solutions within the mathematical system are viable for the actual situation.

**Conclusion:**

**Option C** accurately describes the scenario:  
**"Part of the solution region includes a negative width; therefore, not all solutions are viable for the given situation."**

**Answer:**  
**C. Part of the solution region includes a negative width; therefore, not all solutions are viable for the given situation.**
**C. Part of the solution region includes a negative width; therefore, not all solutions are viable for the given situation.**

Real Tutor Solution

Answer

Solution

Extra Insights

Related Questions

Latest Algebra Questions