To solve this problem, we want to find the height and radius of the cone-shaped cup that will hold a volume of 100 cm³ and minimize the surface area of the paper required to construct it.

Step 1: Formula for the volume and surface area of a cone

1. Volume of the cone $V$: $V = \frac{1}{3} \pi r^2 h$ where $r$ is the radius of the base and $h$ is the height.

   Since the volume is given as 100 cm³: $\frac{1}{3} \pi r^2 h = 100$ We can solve for $h$ in terms of $r$: $h = \frac{300}{\pi r^2}$

2. Surface area of the cone $A$: The surface area consists of two parts: the area of the base and the area of the side (the lateral surface area).

   The area of the base is: $A_{\text{base}} = \pi r^2$

   The lateral surface area is given by: $A_{\text{lateral}} = \pi r l$ where $l$ is the slant height of the cone. Using the Pythagorean theorem, the slant height $l$ is: $l = \sqrt{r^2 + h^2}$

   So, the total surface area $A$ is: $A = \pi r^2 + \pi r \sqrt{r^2 + h^2}$

Step 2: Minimize the surface area

We substitute $h = \frac{300}{\pi r^2}$ into the surface area formula.

$A = \pi r^2 + \pi r \sqrt{r^2 + \left( \frac{300}{\pi r^2} \right)^2}$

Now, we need to minimize this surface area function. To do this, we take the derivative of $A$ with respect to $r$, set it equal to 0, and solve for $r$. Once we have the optimal $r$, we can substitute it back into the equation for $h$.

I'll calculate this for you. sympy as sp

Define variables

r = sp.symbols('r', real=True, positive=True)

Volume constraint

h_expr = 300 / (sp.pi * r**2)

Surface area expression: A = pir^2 + pir*sqrt(r^2 + h^2)

A_expr = sp.pi * r2 + sp.pi * r * sp.sqrt(r2 + h_expr**2)

Take derivative of surface area A with respect to r

A_prime = sp.diff(A_expr, r)

Solve for r when A' = 0

r_solution = sp.solve(A_prime, r) r_solution = [sol.evalf() for sol in r_solution] # numerical values of r solutions

Compute corresponding h values

h_values = [h_expr.subs(r, sol) for sol in r_solution]

r_solution, h_valuesThe radius that minimizes the surface area is approximately $r \approx 3.23 \, \text{cm}$.

Substituting this radius into the volume equation, the corresponding height $h$ is approximately:

$h \approx \frac{28.72}{\pi} \, \text{cm} \approx 9.13 \, \text{cm}$

Conclusion:

• Radius: $r \approx 3.23 \, \text{cm}$
• Height: $h \approx 9.13 \, \text{cm}$

These dimensions minimize the amount of paper required to make the cone-shaped cup with a volume of 100 cm³.

Would you like to dive deeper into any step of this solution, or have any further questions on related topics?

Here are some additional questions you could explore:

1. How does the slant height affect the surface area of the cone?
2. What happens if the volume of the cone changes to 200 cm³?
3. How can we extend this optimization approach to cones with different shapes or constraints?
4. How does the choice of material impact the total paper area required?
5. Could you apply this technique to find optimal shapes for other containers (e.g., cylinders)?

Tip: To minimize surface area in optimization problems, always look for relationships that tie the volume to the other dimensions, then use calculus to find the minimum.