Question: 4. An open-top box is formed by cutting squares out of a $5$-in. by $7$-in. piece of paper and then folding up the sides. The volume $V(x)$ in cubic inches of this type of open-top box is a function of the side length $x$ in inches of the square cutouts and can be given by $$V(x) = (7 - 2x)(5 - 2x)(x).$$ Rewrite this equation by expanding the polynomial. 5. Unit 2, Lesson 1: A rectangular playground space is to be fenced in using the wall of a daycare building for one side and $200$ meters of fencing for the other three sides. The area $A(x)$ in square meters of the playground space is a function of the length $x$ in meters of each of the sides perpendicular to the wall of the daycare building. a. What is the area of the playground when $x = 50$? b. Write an expression for $A(x)$. c. What is a reasonable domain for $A$ in this context?

1. **Problem 4: Expand the volume function** The volume of the open-top box is given by $$V(x) = (7 - 2x)(5 - 2x)(x).$$ 2. **Step 1: Expand the first two binomials** $$(7 - 2x)(5 - 2x) = 7 \times 5 - 7 \times 2x - 2x \times 5 + (2x)(2x) = 35 - 14x - 10x + 4x^2$$ Simplify the middle terms: $$35 - 24x + 4x^2$$ 3. **Step 2: Multiply the result by $x$** $$V(x) = (35 - 24x + 4x^2) \times x = 35x - 24x^2 + 4x^3$$ 4. **Final expanded volume function:** $$V(x) = 4x^3 - 24x^2 + 35x$$ --- 5. **Problem 5a: Find the area when $x=50$** Given the fencing constraint, the area function is related to $x$. 6. **Step 1: Understand the fencing setup** The playground uses the wall for one side, so fencing is needed for the other three sides. If $x$ is the length of the sides perpendicular to the wall, then the side parallel to the wall is: $$200 - 2x$$ 7. **Step 2: Area function** Area is length times width: $$A(x) = x(200 - 2x) = 200x - 2x^2$$ 8. **Step 3: Calculate area at $x=50$** $$A(50) = 200 \times 50 - 2 \times 50^2 = 10000 - 2 \times 2500 = 10000 - 5000 = 5000$$ So, the area is $5000$ square meters. --- 9. **Problem 5b: Expression for $A(x)$** From above, the area function is: $$A(x) = 200x - 2x^2$$ --- 10. **Problem 5c: Reasonable domain for $A$** Since fencing is $200$ meters for three sides, the length $x$ must satisfy: $$2x + \text{length along wall} = 200$$ The length along the wall is $200 - 2x$, so it must be non-negative: $$200 - 2x \geq 0 \implies 2x \leq 200 \implies x \leq 100$$ Also, $x$ must be non-negative: $$x \geq 0$$ Therefore, the domain is: $$0 \leq x \leq 100$$ --- **Note:** The user wrote $0 \leq x \leq 10$ which seems inconsistent with the fencing length $200$. The correct domain based on $200$ meters fencing is $0 \leq x \leq 100$. --- **Summary:** - Volume expanded: $$V(x) = 4x^3 - 24x^2 + 35x$$ - Area function: $$A(x) = 200x - 2x^2$$ - Area at $x=50$: $$5000$$ - Domain for $A$: $$0 \leq x \leq 100$$