Standard soft drink cans hold 12 ounce or 355 milliliters of soda. Assume that soda can must be right circular cylinders. Through trial and error with cans of different sizes, find the dimensions of a 12-ounce can that has the lowest cost for materials.
π r^2 h = 355
r^2 h = 113
the area a=2πrh+2πr^2 which we want to be small
113 is prime, so there will not be integer values. However, using integers to approximate,
r h a/2π(r^2+rh)
3 13 48
4 7 44
5 4 45
6 3 54
Looks like r is near 4
To find the dimensions of a 12-ounce can that has the lowest cost for materials, let's consider the cost of materials as a function of the can's dimensions.
First, we need to define the dimensions of the can. Since a soda can must be a right circular cylinder, it is characterized by its height and radius.
Let's denote the height of the can as h and the radius as r.
Next, we need to determine the cost of materials. The cost of materials can be calculated based on the surface area of the can. The surface area is the sum of the area of the top, bottom, and side of the can.
The area of the top and bottom of the can, which are both circular, can be calculated using the formula for the area of a circle: A = πr².
The area of the side of the can can be calculated as the product of the height and the circumference of the base of the cylinder: A = 2πrh.
Now, we can calculate the total surface area of the can: A = 2πr² + 2πrh.
Since we are interested in finding the dimensions that minimize the cost of materials, we need to minimize the surface area of the can.
We can use calculus to find the minimum value of the surface area. To do this, we differentiate the surface area equation with respect to r and set it equal to zero to find the critical points. Then, we find the second derivative to determine if these critical points correspond to a minimum.
Differentiating the surface area equation with respect to r, we get: dA/dr = 4πr + 2πh = 0.
Setting this equation equal to zero, we find: 4πr = -2πh.
Simplifying, we get: r = -h/2.
Next, we differentiate the equation again to find the second derivative: d²A/dr² = 4π.
Since the second derivative is positive (4π > 0), this means that the critical point r = -h/2 corresponds to a minimum.
Now, we know that r = -h/2. To determine the specific dimensions for the lowest cost can, we need more information. We can use the fact that the can holds 12 ounces or 355 milliliters of soda.
The volume of a can can be calculated using the formula for the volume of a cylinder: V = πr²h.
Substituting r = -h/2, we get: V = π(-h/2)²h.
Simplifying, we get: V = πh³/4.
We know that the volume of the can is 355 milliliters, or 355 cubic centimeters. Since 1 milliliter is equal to 1 cubic centimeter, we can rewrite the volume equation as: 355 = πh³/4.
Solving this equation for h, we find: h = (∛(4 * 355/π)).
Now that we have the value of h, we can calculate the corresponding value of r: r = -h/2.
Once we have the values of h and r, we can compute the cost of materials using the surface area equation: A = 2πr² + 2πrh.
Now that you know how to calculate the dimensions of a 12-ounce can that has the lowest cost for materials, you can use this process to find the specific values for h and r.