1. **State the problem:** We need to find the dimensions of a rectangular container with volume 50 ft³. The container has dimensions: length = $4x + 10$ ft, width = $x$ ft, and height = 1 ft. 2. **Model the volume:** The volume $V$ of a rectangular prism is given by the formula: $$V = \text{length} \times \text{width} \times \text{height}$$ Here, substituting the given dimensions: $$V = (4x + 10) \times x \times 1 = (4x + 10)x$$ 3. **Set up the equation:** We know the volume is 50 ft³, so: $$50 = (4x + 10)x$$ 4. **Simplify the equation:** $$50 = 4x^2 + 10x$$ 5. **Rewrite as a quadratic equation:** $$4x^2 + 10x - 50 = 0$$ 6. **Simplify by dividing all terms by 2:** $$\cancel{2} \times (2x^2 + 5x - 25) = 0 \Rightarrow 2x^2 + 5x - 25 = 0$$ 7. **Solve the quadratic equation using the quadratic formula:** The quadratic formula is: $$x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}$$ where $a=2$, $b=5$, and $c=-25$. Calculate the discriminant: $$\Delta = b^2 - 4ac = 5^2 - 4 \times 2 \times (-25) = 25 + 200 = 225$$ Calculate the roots: $$x = \frac{-5 \pm \sqrt{225}}{2 \times 2} = \frac{-5 \pm 15}{4}$$ 8. **Find the two possible values for $x$:** - $$x = \frac{-5 + 15}{4} = \frac{10}{4} = 2.5$$ - $$x = \frac{-5 - 15}{4} = \frac{-20}{4} = -5$$ 9. **Interpret the solution:** Since $x$ represents a length, it must be positive. So, $x = 2.5$ ft. 10. **Find the other dimensions:** - Width = $x = 2.5$ ft - Length = $4x + 10 = 4(2.5) + 10 = 10 + 10 = 20$ ft - Height = 1 ft (given) **Final answer:** The dimensions of the container are length = 20 ft, width = 2.5 ft, and height = 1 ft.