A soda can is to hold 12 fluid ounces. Suppose that the bottom and top are twice as thick as the sides. Find the dimensions of the can that minimize the amount of material used. (Hint: Instead of minimizing surface area, minimize the cost, which is proportional to the product of the thickness and the area.)
The Correct Answer and Explanation is :
Answer:
The optimal dimensions that minimize the material cost for a soda can holding 12 fluid ounces (≈ 21.66 cubic inches) are:
- Radius (r): ( \sqrt[3]{\frac{21.66}{2\pi}} \approx 1.54 ) inches
- Height (h): ( \frac{21.66}{\pi(1.54)^2} \approx 2.91 ) inches
Explanation:
To minimize material cost, we must consider both surface area and thickness distribution. The thickness of the top and bottom is twice that of the sides, so minimizing cost means optimizing the use of material rather than just minimizing total surface area.
- Volume Constraint
The volume of a cylindrical can is given by:
[
V = \pi r^2 h
]
Given that the can must hold 12 fluid ounces (≈ 21.66 in³), we set up the constraint:
[
\pi r^2 h = 21.66
] - Material Cost Function
- The side surface area is ( 2\pi rh ), with thickness ( t ).
- The top and bottom areas are each ( \pi r^2 ), but their thickness is ( 2t ).
- Total cost is proportional to:
[
C = 2t(2\pi rh) + 2(2t)(\pi r^2)
]
[
C = 4\pi r h t + 4\pi r^2 t
]
[
C = 4\pi r t (h + r)
]
Using the volume constraint to express ( h ) in terms of ( r ):
[
h = \frac{21.66}{\pi r^2}
]
Substituting into ( C ):
[
C = 4\pi r t \left( \frac{21.66}{\pi r^2} + r \right)
]
[
C = 4t(21.66/r + \pi r^2)
]
To minimize cost, differentiate and solve ( dC/dr = 0 ), yielding:
[
r = \sqrt[3]{\frac{21.66}{2\pi}} \approx 1.54 \text{ in}, \quad h = \frac{21.66}{\pi(1.54)^2} \approx 2.91 \text{ in}
]
Thus, a radius of 1.54 inches and a height of 2.91 inches minimizes material cost.