Question: Answer the following. 5. Do you agree that the solution sets of $y < x^2 + x - 20$ and $y > x^2 + x - 20$ is the set of all points on a plane? Justify your answer by graphing the solution set of each on a coordinate plane. 6. Luisa says that the solutions of $y > 2x^2 - 8x + 7$ are also solutions of $y > x^2 - 4x + 3$. Do you agree with Luisa? Justify your answer. 7. A rectangular box is completely filled with dice. Each die has a volume of 1 cm³. The length of the box is 3 cm greater than its width and its height is 5 cm. Suppose the box holds at most 140 dice. What are the possible dimensions of the box? 8. A company decided to increase the size of the box for the packaging of their canned sardines. The length of the original packaging box was 40 cm longer than its width, the height was 12 cm, and the volume was at most 4800 cm³. a. How would you represent the width of the original packaging box? How about the length of the box? b. What expression would represent the volume of the original packaging box? How about the mathematical sentence that would represent its volume? Define the variables used. c. What could be the greatest possible dimensions of the box if each dimension is in whole centimeters? Explain how you arrived at your answer. d. Suppose the length of the new packaging box is still 40 cm longer than its width and the height is 12 cm. What mathematical sentence would represent the volume of the new packaging box? Define the variables used. e. What could be the dimensions of the box? Give the possible dimensions of at least three different boxes.

1. **Problem Statement:** We will answer questions 5 to 8 about quadratic inequalities and box dimensions. --- 2. **Question 5:** Do the solution sets of $y < x^2 + x - 20$ and $y > x^2 + x - 20$ cover the entire plane? - The boundary curve is $y = x^2 + x - 20$. - The inequality $y < x^2 + x - 20$ represents all points below this parabola. - The inequality $y > x^2 + x - 20$ represents all points above this parabola. - Since these two inequalities cover all points except those exactly on the parabola, their union is the entire plane except the parabola itself. - Therefore, the solution sets together cover all points on the plane except the points on the curve $y = x^2 + x - 20$. --- 3. **Question 6:** Are solutions of $y > 2x^2 - 8x + 7$ also solutions of $y > x^2 - 4x + 3$? - Consider the two parabolas: $$f(x) = 2x^2 - 8x + 7$$ $$g(x) = x^2 - 4x + 3$$ - To check if $y > f(x)$ implies $y > g(x)$, check if $f(x) \ge g(x)$ for all $x$. - Compute $f(x) - g(x) = (2x^2 - 8x + 7) - (x^2 - 4x + 3) = x^2 - 4x + 4 = (x - 2)^2 \ge 0$ for all $x$. - Since $f(x) \ge g(x)$, any $y$ greater than $f(x)$ is also greater than $g(x)$. - Therefore, Luisa is correct: solutions of $y > 2x^2 - 8x + 7$ are also solutions of $y > x^2 - 4x + 3$. --- 4. **Question 7:** Box filled with dice. - Let width = $w$ cm. - Length = $w + 3$ cm. - Height = 5 cm. - Volume of box = $w \times (w + 3) \times 5$. - Each die has volume 1 cm³, and box holds at most 140 dice, so: $$w(w + 3)5 \le 140$$ - Simplify: $$5w^2 + 15w \le 140$$ - Divide both sides by 5: $$\cancel{5}w^2 + 3\cancel{5}w \le \frac{140}{5}$$ $$w^2 + 3w \le 28$$ - Rearrange: $$w^2 + 3w - 28 \le 0$$ - Solve quadratic inequality: - Find roots of $w^2 + 3w - 28 = 0$ using quadratic formula: $$w = \frac{-3 \pm \sqrt{3^2 - 4(1)(-28)}}{2} = \frac{-3 \pm \sqrt{9 + 112}}{2} = \frac{-3 \pm \sqrt{121}}{2}$$ $$w = \frac{-3 \pm 11}{2}$$ - Roots: $$w_1 = \frac{-3 - 11}{2} = -7$$ $$w_2 = \frac{-3 + 11}{2} = 4$$ - Since width must be positive, consider $0 < w \le 4$. - **Possible dimensions:** - Width $w$ between 0 and 4 cm. - Length $w + 3$ between 3 and 7 cm. - Height fixed at 5 cm. --- 5. **Question 8:** Packaging box problem. **a. Represent width and length:** - Let width = $w$ cm. - Length = $w + 40$ cm. **b. Volume expression and inequality:** - Height = 12 cm. - Volume $V = w \times (w + 40) \times 12$. - Volume constraint: $$w(w + 40)12 \le 4800$$ - Simplify: $$12w^2 + 480w \le 4800$$ - Divide both sides by 12: $$w^2 + 40w \le 400$$ **c. Greatest possible dimensions (whole cm):** - Rearrange: $$w^2 + 40w - 400 \le 0$$ - Solve quadratic equation $w^2 + 40w - 400 = 0$: $$w = \frac{-40 \pm \sqrt{40^2 - 4(1)(-400)}}{2} = \frac{-40 \pm \sqrt{1600 + 1600}}{2} = \frac{-40 \pm \sqrt{3200}}{2}$$ $$\sqrt{3200} = \sqrt{64 \times 50} = 8\sqrt{50} \approx 56.57$$ $$w = \frac{-40 \pm 56.57}{2}$$ - Roots: $$w_1 = \frac{-40 - 56.57}{2} = -48.29$$ $$w_2 = \frac{-40 + 56.57}{2} = 8.29$$ - Width must be positive and satisfy inequality, so: $$0 < w \le 8.29$$ - Greatest whole number width is 8 cm. - Length = $8 + 40 = 48$ cm. - Height = 12 cm. - Check volume: $$8 \times 48 \times 12 = 4608 \le 4800$$ - So dimensions: width = 8 cm, length = 48 cm, height = 12 cm. **d. New packaging box volume expression:** - Let new width = $w$ cm. - Length = $w + 40$ cm. - Height = 12 cm. - Volume $V = w(w + 40)12$. - Inequality: $$w(w + 40)12 \le V_{new}$$ - $V_{new}$ is the new volume limit (not specified). **e. Possible dimensions of three different boxes:** - Choose widths $w$ such that $w^2 + 40w \le \frac{V_{new}}{12}$. - For example, if $V_{new} = 4800$ (same as original): - $w = 5$ cm, length = 45 cm, volume = $5 \times 45 \times 12 = 2700$ cm³. - $w = 7$ cm, length = 47 cm, volume = $7 \times 47 \times 12 = 3948$ cm³. - $w = 8$ cm, length = 48 cm, volume = $8 \times 48 \times 12 = 4608$ cm³. --- **Final answers:** - 5: The solution sets of $y < x^2 + x - 20$ and $y > x^2 + x - 20$ cover the entire plane except the parabola itself. - 6: Yes, Luisa is correct because $2x^2 - 8x + 7 \ge x^2 - 4x + 3$ for all $x$. - 7: Possible box dimensions satisfy $0 < w \le 4$, length $w+3$, height 5. - 8: a. Width = $w$, length = $w + 40$. b. Volume $= 12w(w + 40)$, with $12w(w + 40) \le 4800$. c. Greatest whole number width is 8 cm, length 48 cm, height 12 cm. d. New box volume $= 12w(w + 40)$ with new volume limit. e. Examples: $(5, 45, 12)$, $(7, 47, 12)$, $(8, 48, 12)$.