2. a geometry teacher distributes sheets of construction paper to her class so that they can create open nets of rectangular prisms. she instructs her students to cut squares from each of the four corners of the construction paper, and then fold up the remaining sides to form a box without a lid. each sheet of construction paper is 24 inches long by 18 inches wide.

a. let s represent the side length of a corner square in inches. write a function v(s) to represent the volume of a box in terms of s.

b. what is the greatest possible integer value of s? explain your reasoning.

c. for what integer value of s is the volume of the box a maximum? state the volume and dimensions of this box.

3. sketch the graph of the cubic function that is the product of the functions shown.

Question

2. a geometry teacher distributes sheets of construction paper to her class so that they can create open nets of rectangular prisms. she instructs her students to cut squares from each of the four corners of the construction paper, and then fold up the remaining sides to form a box without a lid. each sheet of construction paper is 24 inches long by 18 inches wide.

a. let s represent the side length of a corner square in inches. write a function v(s) to represent the volume of a box in terms of s.

b. what is the greatest possible integer value of s? explain your reasoning.

c. for what integer value of s is the volume of the box a maximum? state the volume and dimensions of this box.

3. sketch the graph of the cubic function that is the product of the functions shown.

Accepted Answer

### Part 2a # Explanation: ## Step1: Determine box dimensions When we cut a square of side $ s $ from each corner, the length of the box becomes $ 24 - 2s $ (subtracting $ s $ from both ends), the width becomes $ 18 - 2s $, and the height is $ s $. ## Step2: Volume formula for a rectangular prism The volume $ V $ of a rectangular prism is $ \text{length} \times \text{width} \times \text{height} $. So substituting the dimensions, we get $ V(s)=s(24 - 2s)(18 - 2s) $. # Answer: $ V(s)=s(24 - 2s)(18 - 2s) $ ### Part 2b # Explanation: ## Step1: Analyze non - negative dimensions The length and width of the box must be non - negative. So we have the inequalities: $ 24-2s\geq0$ and $ 18 - 2s\geq0$ For $ 24-2s\geq0$, we solve for $ s $: $ 24\geq2s$ $ s\leq12$ For $ 18 - 2s\geq0$, we solve for $ s $: $ 18\geq2s$ $ s\leq9$ Also, $ s>0$ (since the side length of a square can't be non - positive). ## Step2: Find the greatest integer $ s $ We need to find the largest integer $ s $ such that $ 24 - 2s>0$ and $ 18 - 2s>0$ (because if $ 24-2s = 0$ or $ 18 - 2s=0$, the volume will be $ 0$). From $ 18-2s>0$, we have $ 2s<18$, so $ s < 9$. The greatest integer less than 9 is 8. We can check: If $ s = 8$, length $=24-2\times8=24 - 16 = 8$, width $=18-2\times8=18 - 16 = 2$, height $=8$. Volume $=8\times8\times2 = 128>0$ If $ s=9$, width $=18 - 2\times9=0$, volume $=9\times(24 - 18)\times0 = 0$ # Answer: The greatest possible integer value of $ s $ is 8. Because when $ s = 9$, the width of the box becomes $ 18-2\times9 = 0$ (volume is 0), and for $ s=8$, the length $=24 - 16=8$, width $=18 - 16 = 2$, height $=8$ and volume is positive. ### Part 2c # Explanation: ## Step1: Calculate volume for integer values of $ s $ We can calculate the volume for integer values of $ s $ starting from $ s = 1$ up to $ s = 8$ (from part 2b we know $ s<9$) - For $ s = 1$: $ V(1)=1\times(24 - 2)\times(18 - 2)=1\times22\times16 = 352$ - For $ s = 2$: $ V(2)=2\times(24 - 4)\times(18 - 4)=2\times20\times14 = 560$ - For $ s = 3$: $ V(3)=3\times(24 - 6)\times(18 - 6)=3\times18\times12 = 648$ - For $ s = 4$: $ V(4)=4\times(24 - 8)\times(18 - 8)=4\times16\times10 = 640$ - For $ s = 5$: $ V(5)=5\times(24 - 10)\times(18 - 10)=5\times14\times8 = 560$ - For $ s = 6$: $ V(6)=6\times(24 - 12)\times(18 - 12)=6\times12\times6 = 432$ - For $ s = 7$: $ V(7)=7\times(24 - 14)\times(18 - 14)=7\times10\times4 = 280$ - For $ s = 8$: $ V(8)=8\times(24 - 16)\times(18 - 16)=8\times8\times2 = 128$ ## Step2: Identify maximum volume By comparing the volumes: $ 352,560,648,640,560,432,280,128$, we can see that the maximum volume occurs at $ s = 3$ ## Step3: Find dimensions and volume at $ s = 3$ When $ s = 3$: - Length $=24-2\times3=18$ inches - Width $=18 - 2\times3 = 12$ inches - Height $=3$ inches - Volume $=3\times18\times12=648$ cubic inches # Answer: The integer value of $ s $ for which the volume is maximum is $ 3$. The volume of the box is $ 648$ cubic inches. The dimensions of the box are length $ = 18$ inches, width $=12$ inches and height $ = 3$ inches. ### Part 3 # Explanation: ## Step1: Analyze the function $ f(x)=(x - 1)(x - 1)(x - 4)=(x - 1)^2(x - 4)$ - **Roots**: The roots of the function are found by setting $ f(x)=0$. So $ (x - 1)^2(x - 4)=0$. This gives us a root at $ x = 1$ (with multiplicity 2) and a root at $ x = 4$ (with multiplicity 1). - **End - behavior**: For a cubic function of the form $ ax^3+bx^2+cx + d$, as $ x ightarrow\infty$, if the leading coefficient $ a>0$, $ y ightarrow\infty$ and as $ x ightarrow-\infty$, $ y ightarrow-\infty$. Let's expand $ (x - 1)^2(x - 4)=(x^{2}-2x + 1)(x - 4)=x^{3}-4x^{2}-2x^{2}+8x+x - 4=x^{3}-6x^{2}+9x - 4$. The leading coefficient $ a = 1>0$, so as $ x ightarrow\infty$, $ f(x) ightarrow\infty$ and as $ x ightarrow-\infty$, $ f(x) ightarrow-\infty$. - **Behavior at roots**: - At $ x = 1$ (multiplicity 2), the graph touches the $ x$ - axis and turns around (since the multiplicity is even). - At $ x = 4$ (multiplicity 1), the graph crosses the $ x$ - axis (since the multiplicity is odd). - **Key points**: - When $ x = 0$, $ f(0)=(0 - 1)^2(0 - 4)=1\times(-4)=-4$ - When $ x = 2$, $ f(2)=(2 - 1)^2(2 - 4)=1\times(-2)=-2$ - When $ x = 3$, $ f(3)=(3 - 1)^2(3 - 4)=4\times(-1)=-4$ ## Step2: Sketching the graph 1. Plot the roots: $ (1,0)$ (touch and turn) and $ (4,0)$ (cross). 2. Plot the $ y$ - intercept $ (0,-4)$. 3. Plot the points $ (2,-2)$ and $ (3,-4)$. 4. Use the end - behavior and the behavior at the roots to draw the graph. The graph comes from the bottom left (since as $ x ightarrow-\infty$, $ y ightarrow-\infty$), approaches the $ x$ - axis at $ x = 1$, touches it and turns around, then goes down to pass through the points $ (2,-2)$, $ (3,-4)$ and then crosses the $ x$ - axis at $ x = 4$ and goes up to the top right (since as $ x ightarrow\infty$, $ y ightarrow\infty$). # Answer: To sketch the graph of $ y=(x - 1)^2(x - 4)$: - Plot the root at $ x = 1$ (the graph touches the $ x$ - axis here) and $ x = 4$ (the graph crosses the $ x$ - axis here). - Plot the $ y$ - intercept $ (0,-4)$, and the points $ (2,-2)$ and $ (3,-4)$. - Draw a smooth curve that starts from the bottom left, touches the $ x$ - axis at $ (1,0)$, goes down to pass through the plotted points and then crosses the $ x$ - axis at $ (4,0)$ and goes to the top right, consistent with the end - behavior and the multiplicity of the roots.

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QUESTION IMAGE

Question

Explanation:

Part 2a

Step1: Determine box dimensions

Step2: Volume formula for a rectangular prism

Step1: Analyze non - negative dimensions

Step2: Find the greatest integer \( s \)

Step1: Calculate volume for integer values of \( s \)

Step2: Identify maximum volume

Step3: Find dimensions and volume at \( s = 3\)

Answer:

Part 2b