To solve this problem, we need to minimize the amount of material used to make an open-top cylindrical can with a fixed volume of 390 cubic centimeters. Let’s break down the solution step by step.

Step 1: Define the Variables and Formulas

1. Let $r$ be the radius of the base of the cylinder (in cm).
2. Let $h$ be the height of the cylinder (in cm).

The volume $V$ of a cylinder is given by: $V = \pi r^2 h$ Since we know the volume is 390 cm³, we can write: $\pi r^2 h = 390$

Step 2: Set Up the Surface Area Formula

Since the can is open at the top, the surface area $S$ of the cylinder consists of:

• The base area: $\pi r^2$
• The side area (lateral surface): $2 \pi r h$

Thus, the surface area $S$ is: $S = \pi r^2 + 2 \pi r h$

Step 3: Express $h$ in Terms of $r$

Using the volume formula $\pi r^2 h = 390$, we can solve for $h$ in terms of $r$: $h = \frac{390}{\pi r^2}$

Step 4: Substitute $h$ into the Surface Area Formula

Now substitute $h = \frac{390}{\pi r^2}$ into the surface area formula: $S = \pi r^2 + 2 \pi r \left(\frac{390}{\pi r^2}\right)$

Simplify this expression: $S = \pi r^2 + \frac{780}{r}$

Step 5: Minimize the Surface Area

To find the minimum surface area, take the derivative of $S$ with respect to $r$ and set it equal to zero: $\frac{dS}{dr} = 2 \pi r - \frac{780}{r^2}$

Set the derivative equal to zero: $2 \pi r - \frac{780}{r^2} = 0$

Rearrange to isolate $r$: $2 \pi r^3 = 780$

$r^3 = \frac{780}{2 \pi}$

$r^3 = \frac{390}{\pi}$

Now, take the cube root of both sides: $r = \left(\frac{390}{\pi}\right)^{\frac{1}{3}}$

Step 6: Find $h$

Now that we have $r$, substitute it back into the expression for $h$: $h = \frac{390}{\pi r^2} = \frac{390}{\pi \left(\frac{390}{\pi}\right)^{\frac{2}{3}}}$

Simplify the expression for $h$: $h = \left(\frac{390}{\pi}\right)^{\frac{1}{3}}$

Final Answer

Thus, the radius and height that minimize the amount of material needed are: $\text{Radius} = \left(\frac{390}{\pi}\right)^{\frac{1}{3}}$ $\text{Height} = \left(\frac{390}{\pi}\right)^{\frac{1}{3}}$

Would you like a detailed step-by-step breakdown of any part of this solution?

Here are some related questions to consider:

1. How does changing the volume affect the optimal radius and height?
2. What would the solution look like if the can was closed on top?
3. Can we derive a general formula for any given volume for an open-top cylindrical can?
4. Why does setting the derivative to zero help find the minimum surface area?
5. How does the surface area formula change if the shape of the can changes?

Tip: When dealing with optimization problems involving shapes, always express one variable in terms of another to reduce the number of variables, simplifying differentiation.