Question: 3. An artist is going to enlarge one of their pieces for a public art installation. The largest area that can be used is 40 square feet. The proportions of the piece must remain the same. The original piece has proportions as shown in the picture here: ... x + 1 2x Find the dimensions of the new piece, including the correct units. Explain your steps and how you were able to determine which solution was the correct one.

1. **State the problem:** We have a rectangle with width $$2x$$ and height $$x+1$$. The artist wants to enlarge this piece so that the area is at most 40 square feet, keeping the proportions the same. 2. **Write the formula for area:** The area $$A$$ of a rectangle is given by $$A = \text{width} \times \text{height}$$. 3. **Express the area in terms of $$x$$:** $$A = 2x \times (x+1) = 2x(x+1) = 2x^2 + 2x$$ 4. **Set the area equal to the maximum allowed area:** $$2x^2 + 2x = 40$$ 5. **Simplify the equation:** Divide both sides by 2: $$\cancel{2}x^2 + \cancel{2}x = \frac{40}{2}$$ $$x^2 + x = 20$$ 6. **Rewrite as a quadratic equation:** $$x^2 + x - 20 = 0$$ 7. **Solve the quadratic equation using the quadratic formula:** The quadratic formula is $$x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}$$ where $$a=1$$, $$b=1$$, and $$c=-20$$. Calculate the discriminant: $$\Delta = b^2 - 4ac = 1^2 - 4(1)(-20) = 1 + 80 = 81$$ Calculate the roots: $$x = \frac{-1 \pm \sqrt{81}}{2} = \frac{-1 \pm 9}{2}$$ 8. **Find the two possible solutions:** $$x_1 = \frac{-1 + 9}{2} = \frac{8}{2} = 4$$ $$x_2 = \frac{-1 - 9}{2} = \frac{-10}{2} = -5$$ 9. **Determine the valid solution:** Since $$x$$ represents a length, it must be positive. Therefore, $$x = 4$$ is the valid solution. 10. **Find the dimensions of the new piece:** Width: $$2x = 2 \times 4 = 8$$ feet Height: $$x + 1 = 4 + 1 = 5$$ feet 11. **Verify the area:** $$8 \times 5 = 40$$ square feet, which matches the maximum allowed area. **Final answer:** The dimensions of the new piece are $$8$$ feet by $$5$$ feet.