A cylindrical can is to be made with a fixed volume of 1000 cm³. What dimensions

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Question: A cylindrical can is to be made with a fixed volume of 1000 cm³. What dimensions minimize the surface area? (2022)

Options:

Correct Answer: 5 cm height, 15.87 cm radius

Exam Year: 2022

Solution:

Using the formula for surface area and volume, the optimal dimensions are found to be 5 cm height and approximately 15.87 cm radius.

A cylindrical can is to be made with a fixed volume of 1000 cm³. What dimensions minimize the surface area? (2022)

A cylindrical can is to be made with a fixed volume of 1000 cm³. What dimensions minimize the surface area? (2022)

Step 1: Understand the problem. We need to find the dimensions (height and radius) of a cylindrical can that has a volume of 1000 cm³ and minimizes the surface area.
Step 2: Write down the formulas. The volume V of a cylinder is given by V = πr²h, where r is the radius and h is the height. The surface area A is given by A = 2πr² + 2πrh.
Step 3: Set the volume equal to 1000 cm³. So, we have πr²h = 1000. We can express h in terms of r: h = 1000 / (πr²).
Step 4: Substitute h into the surface area formula. Replace h in A = 2πr² + 2πrh with the expression we found: A = 2πr² + 2πr(1000 / (πr²)).
Step 5: Simplify the surface area formula. This gives us A = 2πr² + 2000/r.
Step 6: To minimize the surface area, take the derivative of A with respect to r and set it to zero. This will help us find the radius that minimizes the surface area.
Step 7: Solve the equation from Step 6 to find the optimal radius r. You will find that r is approximately 15.87 cm.
Step 8: Use the radius to find the height. Substitute r back into the equation h = 1000 / (πr²) to find the height, which will be approximately 5 cm.
Step 9: Conclude that the optimal dimensions for the can are a radius of approximately 15.87 cm and a height of 5 cm.

Optimization – The process of finding the minimum or maximum value of a function, in this case, minimizing the surface area of a cylinder given a fixed volume.
Geometry of Cylinders – Understanding the relationship between the dimensions (radius and height) of a cylinder and its surface area and volume.
Calculus – Utilizing derivatives to find critical points that minimize the surface area function.