The problem you are working on involves finding the height and radius of a cylindrical can that minimize the material needed to hold 890 cm³ of liquid. This can is open at the top, and you need to express your answer using rational exponents and $\pi$ to represent pi.

Step-by-Step Solution:

1. Volume of the Cylinder: Since the cylinder has a base and an open top, the volume $V$ is given by: $V = \pi r^2 h$ where $r$ is the radius and $h$ is the height. The volume is given as 890 cm³, so: $\pi r^2 h = 890$

2. Surface Area of the Cylinder: The surface area $A$ of a cylindrical can open at the top consists of:

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

   Thus, the total surface area to minimize is: $A = \pi r^2 + 2\pi r h$

3. Express $h$ in Terms of $r$: From the volume formula, solve for $h$: $h = \frac{890}{\pi r^2}$

4. Substitute $h$ into the Surface Area Formula: Now, substitute the expression for $h$ into the surface area formula: $A = \pi r^2 + 2\pi r \left(\frac{890}{\pi r^2}\right)$ Simplifying this expression: $A = \pi r^2 + \frac{1780}{r}$

5. Minimize the Surface Area: To minimize the surface area, take the derivative of $A$ with respect to $r$ and set it equal to zero: $\frac{dA}{dr} = 2\pi r - \frac{1780}{r^2}$ Set the derivative equal to zero: $2\pi r = \frac{1780}{r^2}$ Multiply both sides by $r^2$: $2\pi r^3 = 1780$ Solve for $r$: $r^3 = \frac{1780}{2\pi}$ $r = \left(\frac{890}{\pi}\right)^{1/3}$

6. Find the Height $h$: Now, use the expression for $r$ to find $h$. From the volume equation: $h = \frac{890}{\pi r^2}$ Substitute $r = \left(\frac{890}{\pi}\right)^{1/3}$ into this expression to get the height.

It seems that the radius input was correct in your submission, but the height was incorrect. Recalculating the height using the steps above should correct the error.

Would you like me to compute the exact values for both the radius and height?

Additional Related Questions:

1. Can you show how to compute the derivative for more complex surface area functions?
2. What are other ways to minimize surface area for different shapes like cones or spheres?
3. How does the open-top condition affect the surface area compared to a closed can?
4. How would you modify the problem if the volume was larger or smaller?
5. Could the same optimization be done using a Lagrange multiplier?

Tip: When optimizing with a constraint (like volume), always express one variable in terms of the other before differentiating.