To find the dimensions of the cylindrical cup-of-soup package that will minimize the production cost, we can follow these steps, using the given information:

Step 1: Set up the equations for volume and surface area.

We know the volume of a cylinder is given by the formula:

$V = \pi r^2 h$

where:

• $V = 600 \, \text{cm}^3$ (the volume of the soup),
• $r$ is the radius of the base,
• $h$ is the height of the cylinder.

We also need to find the surface area of the cylinder, which consists of:

1. The area of the sides (lateral surface area), which is:

$A_{\text{sides}} = 2\pi rh$

2. The area of the bottom (made of styrofoam), which is:

$A_{\text{bottom}} = \pi r^2$

3. The area of the top (made of glued paper), which is:

$A_{\text{top}} = \pi r^2$

Step 2: Express the cost function.

The cost function depends on the surface areas and the material costs:

• Styrofoam costs 0.03 cents per square centimeter (used for the sides and bottom),
• Glued paper costs 0.07 cents per square centimeter (used for the top).

Thus, the total cost $C$ is:

$C = 0.03(A_{\text{sides}} + A_{\text{bottom}}) + 0.07(A_{\text{top}})$

Substitute the areas into the cost function:

$C = 0.03(2\pi rh + \pi r^2) + 0.07(\pi r^2)$

Simplifying:

$C = 0.03(2\pi rh) + 0.03(\pi r^2) + 0.07(\pi r^2)$ $C = 0.06\pi rh + (0.03 + 0.07)\pi r^2$ $C = 0.06\pi rh + 0.10\pi r^2$

Step 3: Use the volume constraint to eliminate $h$.

From the volume formula $V = \pi r^2 h = 600$, solve for $h$:

$h = \frac{600}{\pi r^2}$

Substitute this expression for $h$ into the cost function:

$C = 0.06\pi r\left(\frac{600}{\pi r^2}\right) + 0.10\pi r^2$

Simplifying:

$C = 0.06 \times \frac{600}{r} + 0.10\pi r^2$ $C = \frac{36}{r} + 0.10\pi r^2$

Step 4: Minimize the cost function.

To find the value of $r$ that minimizes the cost, take the derivative of $C$ with respect to $r$ and set it equal to zero:

$\frac{dC}{dr} = -\frac{36}{r^2} + 0.20\pi r = 0$

Solving for $r$:

$\frac{36}{r^2} = 0.20\pi r$

Multiply both sides by $r^2$ to eliminate the denominator:

$36 = 0.20\pi r^3$

Solve for $r^3$:

$r^3 = \frac{36}{0.20\pi} = \frac{36}{0.628} \approx 57.32$

Take the cube root:

$r \approx \sqrt[3]{57.32} \approx 3.85 \, \text{cm}$

Step 5: Calculate the height $h$.

Using the volume constraint $V = \pi r^2 h = 600$, solve for $h$:

$h = \frac{600}{\pi (3.85)^2} \approx \frac{600}{46.61\pi} \approx 5.12 \, \text{cm}$

Step 6: Calculate the minimum cost.

Substitute $r \approx 3.85 \, \text{cm}$ and $h \approx 5.12 \, \text{cm}$ into the cost function:

$C = \frac{36}{3.85} + 0.10\pi (3.85)^2$ $C \approx 9.35 + 0.10 \times 46.61\pi \approx 9.35 + 14.66 \approx 24.01 \, \text{cents}$

Final Answer:

• Radius: $3.85 \, \text{cm}$
• Height: $5.12 \, \text{cm}$
• Minimum cost: $24.01 \, \text{cents}$

Would you like further details or clarifications?

Related Questions:

1. What is the relationship between the height and radius in a cylinder for a given volume?
2. How does the cost of materials affect the optimization of dimensions?
3. How would the problem change if the top and bottom were made of the same material?
4. What if the cost of the styrofoam was more expensive than the glued paper?
5. Can this method be applied to other geometric shapes to minimize production costs?

Tip:

When optimizing, always express one variable in terms of the others using constraints (like volume) to reduce the complexity of the problem.