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each question write the answer based on the marking scheme 

To address the request of writing answers based on the marking scheme, I will follow the provided solutions for the questions and format the answers accordingly. Below is a breakdown for each question answered as per the scheme found in your file.

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### **Question 1a: Express \( h \) in terms of \( r \).**

**Answer:**
We are given the volume of a cylindrical can, \( V = \pi r^2 h \), and the volume is \( 20 \, \text{cm}^3 \).
1. Start with the equation:  
   \[ \pi r^2 h = 20 \]  
   (A1 mark for the correct volume formula).
2. Solve for \( h \):  
   \[ h = \frac{20}{r^2} \]  
   (A1 mark for correctly solving for \( h \)).

**[2 marks]**

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### **Question 1b: Show that \( C = 20\pi r^2 + \frac{320\pi}{r} \).**

**Answer:**
1. The cost for the base and top of the can is based on the area of two circles:  
   \[ 2 \times \pi r^2 \times 10 = 20\pi r^2 \]  
   (A1 mark for correct formula for the cost of two circles).
2. The cost for the curved side is based on the circumference and height:  
   \[ 2\pi r \times h \times 8 = 16\pi r h \]  
   (A1 mark for correct formula for the curved side).
3. Substitute \( h = \frac{20}{r^2} \) into the expression for the side cost:  
   \[ 16\pi r \times \frac{20}{r^2} = \frac{320\pi}{r} \]  
   (A1 mark for correct expression for the curved side in terms of \( r \)).
4. The total cost is:  
   \[ C = 20\pi r^2 + \frac{320\pi}{r} \]  
   (AG – answer given mark).

**[4 marks]**

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### **Question 1c: Find the minimum value of \( C \) in terms of \( \pi \).**

**Answer:**
1. Differentiate \( C = 20\pi r^2 + \frac{320\pi}{r} \) with respect to \( r \):  
   \[ C' = 40\pi r - \frac{320\pi}{r^2} \]  
   (A1 mark for correct differentiation).
2. Set \( C' = 0 \) to find the critical points:  
   \[ 40\pi r = \frac{320\pi}{r^2} \]  
   (A1 mark for setting the equation to zero).
3. Solve for \( r \):  
   \[ r^3 = 8 \quad \Rightarrow \quad r = 2 \, \text{cm} \]  
   (A1 mark for correct working).
4. Substitute \( r = 2 \) into \( C = 20\pi r^2 + \frac{320\pi}{r} \):  
   \[ C = 20\pi (2^2) + \frac{320\pi}{2} = 80\pi + 160\pi = 240\pi \]  
   (M1 for attempting substitution, A1 for correct working).
5. The minimum value of \( C \) is \( 240\pi \) cents.

**[9 marks]**

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If you want a detailed explanation for any of these steps or solutions, feel free to ask! Here are some relative questions to expand on this:

1. How does one differentiate cost functions involving fractions?
2. What are the conditions for finding minimum or maximum values of cost functions?
3. Can you explain how to optimize the surface area of complex shapes?
4. How does substituting expressions help in reducing terms during optimization?
5. What is the significance of cylindrical optimization in practical design?

**Tip:** Always double-check your differentiation before finding critical points in optimization problems to avoid missing crucial steps.

each question write the answer based on the marking scheme

Learn how to solve an optimization problem involving a cylindrical can with a fixed volume. This solution covers key concepts in cylindrical geometry, differentiation, and cost function optimization. Perfect for IB Diploma students.

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