To solve the problem of minimizing the cost of the metal for manufacturing a cylindrical can that holds 1000 cm³ of oil, we need to minimize the surface area of the can for a given volume.

Variables and Equations:

Let:

• $r$ be the radius of the base of the cylinder.
• $h$ be the height of the cylinder.
• $V = 1000 \, \text{cm}^3$ be the volume of the can.

Step 1: Volume of the Cylinder

The volume of a cylinder is given by: $V = \pi r^2 h$ Given $V = 1000 \, \text{cm}^3$, we can express $h$ in terms of $r$: $1000 = \pi r^2 h$ $h = \frac{1000}{\pi r^2}$

Step 2: Surface Area of the Cylinder

The surface area $S$ of the cylinder, which we want to minimize, consists of two parts: the area of the sides and the area of the two circular bases. It is given by: $S = 2\pi r h + 2\pi r^2$ Substitute $h = \frac{1000}{\pi r^2}$ into this equation: $S = 2\pi r \left( \frac{1000}{\pi r^2} \right) + 2\pi r^2$ Simplify: $S = \frac{2000}{r} + 2\pi r^2$

Step 3: Minimize the Surface Area

Now, we need to minimize $S(r) = \frac{2000}{r} + 2\pi r^2$. To find the critical points, take the derivative of $S$ with respect to $r$, and set it equal to zero.

$\frac{dS}{dr} = -\frac{2000}{r^2} + 4\pi r$ Set $\frac{dS}{dr} = 0$: $-\frac{2000}{r^2} + 4\pi r = 0$ Solve for $r$: $\frac{2000}{r^2} = 4\pi r$ Multiply both sides by $r^2$: $2000 = 4\pi r^3$ $r^3 = \frac{2000}{4\pi}$ $r^3 = \frac{500}{\pi}$ $r = \left( \frac{500}{\pi} \right)^{1/3}$

This gives the optimal radius $r$. Now, calculate the height $h$ using the formula $h = \frac{1000}{\pi r^2}$.

Step 4: Approximate Values

Using $\pi \approx 3.1416$: $r = \left( \frac{500}{3.1416} \right)^{1/3} \approx 5.419 \, \text{cm}$ Now, calculate $h$: $h = \frac{1000}{\pi (5.419)^2} \approx 10.837 \, \text{cm}$

Conclusion:

• The radius of the can should be approximately 5.42 cm.
• The height of the can should be approximately 10.84 cm.

These dimensions minimize the surface area, and therefore minimize the cost of the metal used to manufacture the can.

Would you like further details or have any questions about this solution?

Related Questions:

1. How do we confirm that this is a minimum for the surface area?
2. Can this method be applied to different volumes of oil?
3. What would happen to the height if the radius is increased slightly?
4. How does the derivative help us find the optimal dimensions?
5. What role does the surface area play in manufacturing cost estimation?

Tip: Always double-check units when solving real-world optimization problems. Miscalculations in units can lead to incorrect results.