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Consider the can in the previous exercise. Suppose the tin for the top and bottom of the can cost 8 cents per square inch, while the tin for the remainder of the can costs 1 cent per square inch. What should the diameter of the can be if the cost of the can is to be minimized.

$$d=\frac{3}{1 / 3}, h=\frac{24}{1 / 3}$$

Calculus 1 / AB

Chapter 3

Applications of the Derivative

Section 4

Applications I - Geometric Optimization Problems

Derivatives

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University of Michigan - Ann Arbor

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Lectures

04:40

In mathematics, a derivative is a measure of how a function changes as its input changes. Loosely speaking, a derivative can be thought of as how much one quantity is changing in response to changes in some other quantity; for example, the derivative of the position of a moving object with respect to time is the object's velocity. The concept of a derivative developed as a way to measure the steepness of a curve; the concept was ultimately generalized and now "derivative" is often used to refer to the relationship between two variables, independent and dependent, and to various related notions, such as the differential.

30:01

In mathematics, the derivative of a function of a real variable measures the sensitivity to change of the function value (the rate of change of the value of the function). If the derivative of a function at a chosen input value equals a constant value, the function is said to be a constant function. In this case the derivative itself is the constant of the function, and is called the constant of integration.

03:10

A company manufactures cyl…

06:06

A box with a volume of 8 $…

04:52

A cylindrical can without …

I'm gonna do a problem 15, which I think somebody else did. Um and uh On the website but um it's related to problems 16, so I'm just going to redo it. Um and then two problems 16 in the same video. Um So yeah, well asked about um to optimize the size of an aluminum can, basically a circular cylinder top and bottom, you know, being closed. So the volume is pi r squared H. And we want that volume to be 55 54 cubic inches. The surface area is two pi R squared, which is the pi r squared is the area of the top and the bottom. So we have to and then the sides are the circumference times the height. All right, we can solve for h from here at our constraint And plug it into our surface area and we get the surface area is too over our times 54 -9 R cubed. And so we want to minimize the surface area. So we take the derivative said article are one Saturday equals zero and we get our one equals three divided by the cube root of pi and that's about zero point 2.05". H is actually twice this is actually winds up being six um six divided by the cube root of pi and that's 4.1 inches. So we have a total surface area of 79" cube. And so um if you think about this, the diameter of the of the can is the same as the height. So it's it's very kind of squat, you know, and that that's definitely not what we see in cans, canned goods at the store. Um and I'm not sure exactly why that is, but it's maybe that, you know, there's lots of other optimization problem parts of this, but maybe it's too hard to hold if you have that. You know, if you have, if you have this can because you want this volume um you know, that's kind of hard to hold with a four inch um four inch diameter. And it reminded me of a of a of an episode of bird girl on adult swim where they had this um they were selling this um soda can papa papa. That basically was like a looks like an optimized can basically, it seemed like there the diameter was roughly the same as there of the height. And then in the next problem they asked us to consider cost in this case, so we want the same volume, but now we want the cost. And they said the cost of the top it is eight cents per square inch. And while the 10 for the remainder of the can, the top of the bottom is eight cents per square inch and the rest of the can wind up being one cent per square inch. So our cost winds up being um 16 pi r squared plus two pi R. H. And that would be incense. So now we just solve solve the, solve this for age, plug it into here and we get this is our cost given our volume constraint, take the derivative respect are said are equal to R1 and set that to zero. And now we get that the radius is half of what it was in the previous problem. And the, so the diameter, so we get a radius of two Uh 1.02" and the diameter is 0.2.05". Yeah and let's see here the height then is 24 over pie of a cube root of pi which that is 16 .4". So now we got pretty much a very skinny can. Um and so I don't know if that's 16" is really tall. So it's definitely not not the size of it. Maybe, I don't know, not even like a red bull or like a monster energy drink can or something. This is a really tall can. And why? Why? It's really tall is because the cost of the sidewalls is cheap And so the total cost is in a dollar on $1.12 cents

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