A container for a soft drink is in the form of a right...

Name: Problem 14, Chapter 3: Calculus Early Transcendentals
Uploaded: 2020-12-09T16:32:45-08:00
Duration: 4 min 6 s
Channel: Madi Sousa
Description: A container for a soft drink is in the form of a right circular cylinder. If the container is to have a capacity of 12 fluid ounces (fl oz), find the dimensions of the container that can be constructed with a minimum of material.

A container for a soft drink is in the form of a right circular cylinder. If the container is to have a capacity of 12 fluid ounces (fl oz), find the dimensions of the container that can be constructed with a minimum of material.

Key Concepts

Optimization

Optimization involves finding the best solution, often a minimum or maximum value, under given constraints. In this context, the task is to minimize the surface area, which directly relates to the material used, while maintaining a fixed volume, ejemplifying a constrained optimization problem.

Volume of a Cylinder

The volume of a right circular cylinder is given by the formula V = ?r²h, where r is the radius and h is the height. This fundamental relationship serves as the constraint in the optimization process, ensuring that any solution meets the needed volume requirement.

Surface Area of a Cylinder

The surface area of a cylinder, including both the lateral (side) and the top and bottom areas, is calculated as A = 2?rh + 2?r². Minimizing this surface area function, while adhering to the volume constraint, is central to finding the dimensions that use the least material.

Calculus-based Optimization

Calculus-based optimization involves using techniques such as substitution (to incorporate the volume constraint into the surface area function) and differentiation to find critical points. By setting the derivative equal to zero, one identifies potential minima, and further analysis (like a second derivative test) confirms that the solution indeed minimizes the surface area.

Transcript

00:01 In this problem, we have to understand optimization.

00:05 And when you hear the word optimization, normally in this context, your mind should go, i have to use differentiation.

00:12 So that's what we're going to be using.

00:14 Not only does optimization, pardon me, optimization incorporate differential calculus, but we also need additional information such as a volume equation, which we're going to be using today.

00:27 So let's first understand the equation for volume.

00:31 Volume equals pi r squared times h, where r is our radius and h is our height.

00:37 And we're told that our volume in this case is 12 ounces.

00:42 We're also given the conversion factor to convert ounces into cubic inches.

00:48 So that's the first thing that we're going to do.

00:50 We're going to have 12 times 1 .80469 cubic inches.

00:55 And when we do that, we'll get 21 .656 cubic inches.

01:01 And remember that this 21 and then sum value is still equivalent to our volume.

01:07 So then we can rearrange that to get h by itself.

01:10 And we'll get h as equivalent to 21 .656 over pi r squared, h.

01:18 So that means that our area value is going to be equal to 2 pi r h plus pi r squared.

01:27 And we want to minimize the area.

01:30 Remember, this is all about optimization.

01:33 So we're going to minimize our area.

01:37 So what does that mean for us? that means our area a is going to be equal to 2 pi r times 21 .656 over pi r squared plus pi r squared...

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