Let's solve the problem step by step as requested: --- **1. Unit conversion of fluid ounces to cubic inches** We know: - 1 US fluid ounce = 1.80469 cubic inches So, \[ 16 \text{ fluid ounces} = 16 \times 1.80469 = 28.875 \text{ cubic inches} \] --- **2. Volume equation (Cylinder)** The volume \( V \) of a right circular cylinder is: \[ V = \pi r^2 h \] where: - \( r \) = radius - \( h \) = height Given \( V = 28.875 \) cubic inches: \[ \pi r^2 h = 28.875 \] \[ h = \frac{28.875}{\pi r^2} \] --- **3. Cost function** - **Bottom and top areas:** Each = \( \pi r^2 \) - **Side area:** \( 2\pi r h \) **Costs:** - Bottom: 4 cents/in² - Top: 2 cents/in² - Side: 4 cents/in² So, total cost \( C \) is: \[ C = \text{(Bottom area)} \times 4 + \text{(Top area)} \times 2 + \text{(Side area)} \times 4 \] \[ C = 4(\pi r^2) + 2(\pi r^2) + 4(2\pi r h) \] \[ C = 6\pi r^2 + 8\pi r h \] Substitute \( h \) from above: \[ C = 6\pi r^2 + 8\pi r \left(\frac{28.875}{\pi r^2}\right) \] \[ C = 6\pi r^2 + 8 \times 28.875 \frac{r}{r^2} \] \[ C = 6\pi r^2 + \frac{231}{r} \] --- **4. Derivative of cost function** Let’s differentiate \( C(r) \): \[ C(r) = 6\pi r^2 + \frac{231}{r} \] \[ \frac{dC}{dr} = 12\pi r - \frac{231}{r^2} \] Set derivative to zero for minimum: \[ 12\pi r - \frac{231}{r^2} = 0 \] \[ 12\pi r = \frac{231}{r^2} \] \[ 12\pi r^3 = 231 \] \[ r^3 = \frac{231}{12\pi} \] \[ r = \sqrt[3]{\frac{231}{12\pi}} \] --- **5. Second derivative test** \[ \frac{d^2C}{dr^2} = 12\pi + \frac{462}{r^3} \] Both terms are positive for \( r > 0 \), so \( C(r) \) is minimized at the critical point. --- **6. The radius and height that minimizes cost** **Calculate \( r \):** \[ r = \sqrt[3]{\frac{231}{12\pi}} \] Calculate \( 12\pi \): \[ 12\pi \approx 37.699 \] So, \[ r = \sqrt[3]{\frac{231}{37.699}} = \sqrt[3]{6.127} \] \[ r \approx 1.83 \text{ in} \] **Calculate \( h \):** Recall: \[ h = \frac{28.875}{\pi r^2} \] Substitute \( r \approx 1.83 \): \[ h = \frac{28.875}{\pi (1.83)^2} \] \[ h = \frac{28.875}{\pi \times 3.3489} \] \[ h = \frac{28.875}{10.519} \] \[ h \approx 2.75 \text{ in} \] --- ## **Final Answer:** - The **radius** that minimizes cost is **about 1.83 inches**. - The **height** that minimizes cost is **about 2.75 inches**. --- Let me know if you need a graph or further explanation!

Certainly! Let's go through each step with detailed calculations and explanations. --- ## **Step 1: Convert fluid ounces to cubic inches** Given: - Volume of can = 16 fluid ounces - 1 fluid ounce = 1.80469 cubic inches **Calculation:** \[ \text{Volume} = 16 \times 1.80469 = 28.875 \text{ cubic inches} \] **Result:** \[ V = 28.875 \text{ in}^3 \] --- ## **Step 2: Write the volume equation** For a right circular cylinder: \[ V = \pi r^2 h \] Given \( V = 28.875 \) in³: \[ \pi r^2 h = 28.875 \] Solve for \( h \): \[ h = \frac{28.875}{\pi r^2} \] --- ## **Step 3: Write the cost function** Costs: - Bottom area: \( \pi r^2 \), cost per in² = 2 cents - Top area: \( \pi r^2 \), cost per in² = 2 cents - Side area: \( 2 \pi r h \), cost per in² = 4 cents Total cost \( C \): \[ C = \text{(bottom)} + \text{(top)} + \text{(side)} \] \[ C = 2 \times \pi r^2 + 2 \times \pi r^2 + 4 \times 2 \pi r h \] \[ C = 2\pi r^2 + 2\pi r^2 + 8 \pi r h \] \[ C = 4 \pi r^2 + 8 \pi r h \] **Note:** The problem states: - Side costs 4 cents/in² - Bottom costs 2 cents/in² - Top costs 2 cents/in² **But from the problem statement:** > The material costs of the side and bottom is 4 cents per square inch and the top is 2 cents per square inch. *Assuming:* - Side: 4 cents/in² - Bottom: 4 cents/in² - Top: 2 cents/in² **Recalculate accordingly:** \[ C = 4 \times \pi r^2 + 4 \times \pi r^2 + 2 \times \pi r^2 + 4 \times 2 \pi r h \] But the initial understanding was: - Side: 4 cents/in² - Bottom: 4 cents/in² - Top: 2 cents/in² Total cost: \[ C = (4 \pi r^2) + (4 \pi r^2) + (2 \pi r^2) + \text{side cost} \] Actually, to avoid confusion, let's clarify the costs: **Given:** - Side cost = 4 cents/in² - Bottom cost = 4 cents/in² - Top cost = 2 cents/in² Total cost: \[ C = \text{bottom} + \text{top} + \text{side} \] \[ C = 4 \pi r^2 + 2 \pi r^2 + 4 \times 2 \pi r h \] \[ C = (4 \pi r^2 + 2 \pi r^2) + 8 \pi r h \] \[ C = 6 \pi r^2 + 8 \pi r h \] **Final cost function:** \[ \boxed{ C(r) = 6 \pi r^2 + 8 \pi r h } \] --- ## **Step 4: Express \( C \) solely in terms of \( r \)** Recall: \[ h = \frac{28.875}{\pi r^2} \] Substitute into the cost function: \[ C(r) = 6 \pi r^2 + 8 \pi r \times \frac{28.875}{\pi r^2} \] Simplify: \[ C(r) = 6 \pi r^2 + 8 \times 28.875 \times \frac{r}{r^2} \] \[ C(r) = 6 \pi r^2 + 8 \times 28.875 \times \frac{1}{r} \] Calculate \( 8 \times 28.875 \): \[ 8 \times 28.875 = 231 \] So, \[ C(r) = 6 \pi r^2 + \frac{231}{r} \] --- ## **Step 5: Find the critical point by differentiating \( C(r) \)** Differentiate: \[ C(r) = 6 \pi r^2 + \frac{231}{r} \] \[ \frac{dC}{dr} = 12 \pi r - \frac{231}{r^2} \] Set derivative to zero for minimum: \[ 12 \pi r - \frac{231}{r^2} = 0 \] \[ 12 \pi r = \frac{231}{r^2} \] Multiply both sides by \( r^2 \): \[ 12 \pi r^3 = 231 \] Solve for \( r^3 \): \[ r^3 = \frac{231}{12 \pi} \] Calculate denominator: \[ 12 \pi \approx 12 \times 3.1416 = 37.699 \] Thus: \[ r^3 \approx \frac{231}{37.699} \approx 6.126 \] Calculate \( r \): \[ r = \sqrt[3]{6.126} \approx 1.82 \text{ inches} \] --- ## **Step 6: Calculate the height \( h \)** Recall: \[ h = \frac{28.875}{\pi r^2} \] Calculate \( r^2 \): \[ r^2 = (1.82)^2 \approx 3.312 \] Calculate denominator: \[ \pi r^2 \approx 3.1416 \times 3.312 \approx 10.41 \] Finally: \[ h = \frac{28.875}{10.41} \approx 2.78 \text{ inches} \] --- ## **Final Results:** \[ \boxed{ \text{Radius } r \approx \mathbf{1.82 \text{ inches}} \\ \text{Height } h \approx \mathbf{2.78 \text{ inches}} } \] These dimensions minimize the cost of producing the can. --- **Summary:** - **Radius:** approximately **1.82 inches** - **Height:** approximately **2.78 inches** Let me know if you'd like a graph of the cost function or further clarification!