1. **Problem Statement:** We have a 30-inch by 16-inch piece of cardboard. Squares of side length $x$ are cut from each corner, and the sides are folded up to form an open box. We want to find the volume $V$ of this box as a function of $x$, determine the domain and range of $V$, and interpret the graph of $V$. 2. **Formula for Volume:** The volume $V$ of the box is given by the product of its length, width, and height after cutting and folding. - Original length = 30 inches - Original width = 16 inches - Height after folding = $x$ After cutting squares of side $x$, the new length and width are: - Length = $30 - 2x$ - Width = $16 - 2x$ So, the volume function is: $$V(x) = x(30 - 2x)(16 - 2x)$$ 3. **Domain of $V$:** Since $x$ represents the side length of the squares cut out, it must be positive and small enough so that the length and width remain positive: - $x > 0$ - $30 - 2x > 0 \Rightarrow x < 15$ - $16 - 2x > 0 \Rightarrow x < 8$ The domain is the intersection of these conditions: $$0 < x < 8$$ 4. **Range of $V$ (from graph):** The graph of $V$ shows volume values starting at 0 when $x=0$, increasing to a maximum volume, then decreasing back to 0 at $x=8$. The estimated range is: $$0 < V(x) \leq V_{max}$$ where $V_{max}$ is the maximum volume observed on the graph. 5. **Interpretation of the Graph:** The graph tells us that as the size of the cut squares $x$ increases from 0, the volume of the box increases until it reaches a maximum. Beyond this point, increasing $x$ reduces the volume because the base dimensions shrink too much. At $x=8$, the volume returns to zero because the width becomes zero. **Final answers:** - Volume formula: $$V(x) = x(30 - 2x)(16 - 2x)$$ - Domain: $$0 < x < 8$$ - Range: $$0 < V(x) \leq V_{max}$$ (from graph) This explains how to find and interpret the volume function for the open box.