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The volume of a cylinder is a key concept when dealing with optimization problems in calculus. It's the amount of space inside the cylindrical shape and can be calculated using the formula: \[ V = \pi r^2 h \]. Here, \( V \) represents the volume, \( \pi \) is a mathematical constant approximated as 3.14159, \( r \) is the radius of the circular base, and \( h \) is the height of the cylinder.
To solve the exercise problem, we use the given volume of the orange juice can, which is \( 48 \pi \, \text{cm}^3 \). This helps establish a relationship between the radius and height of the cylinder by rearranging the formula to \[ h = \frac{48}{r^2} \]. This equation allows us to express one dimension in terms of the other, making it easier to work with when minimizing costs.

When discussing the surface area of a cylinder, we refer to both the top and bottom circles and the side or lateral surface.
The total surface area can be expressed as the sum of these components:

• Top and bottom (two circles): \( 2\pi r^2 \)
• Side (lateral surface): \( 2\pi rh \)

The top and bottom circles' area, \( 2\pi r^2 \), is significant because these parts of the can are made of metal, which costs more. The lateral surface \( 2\pi rh \) is covered with cardboard.
Understanding the surface area formula is essential in calculating the cost, as it directly influences the material required and its associated expense.

Cost function minimization is a technique in calculus where we seek to minimize the cost associated with certain variables—in this case, the materials needed to produce the can.
The cost function provides a way to quantify expenses based on dimensions. Here, the cost function is structured as:

• Cost for metal ends: \( 6\pi r^2 \)
• Cost for cardboard sides: \( 2\pi rh \)

Combining these, we create a total cost function: \[ C = 6\pi r^2 + 2\pi rh \].By substituting the expression for \( h \) derived from the volume constraint \( h = \frac{48}{r^2} \), we reformulate the cost function as: \[ C = 6\pi r^2 + \frac{96\pi}{r} \].
Minimizing this function involves using calculus techniques, such as taking derivatives, to find the optimal dimensions that result in the lowest cost.

Critical points are fundamental in calculus as they help determine where functions reach a maximum, minimum, or other turning points.
In optimization tasks like minimizing costs, we find critical points by setting the derivative of the cost function equal to zero. This exercise leads to differentiating \[ \frac{dC}{dr} = 12\pi r - \frac{96\pi}{r^2} \].By solving \( 12\pi r = \frac{96\pi}{r^2} \), we find \( r = 2 \).
Critical points give potential candidates for optimization, but we need to confirm whether these points indeed provide a minimum. We do so through the second derivative test, examining \[ \frac{d^2C}{dr^2} = 12\pi + \frac{192\pi}{r^3} \].The positive value confirms a minimum cost at \( r = 2 \). Thus, critical points and their analysis are crucial in solving optimization problems effectively.