Minimizing Surface Area of a Can

In summary, the conversation discusses the process of finding the radius of a minimized pop can. The equation for surface area is given and the derivative is taken to find critical points. After correcting a mistake in the differentiation, it is determined that the only minimum turning point is at r=0, while the other root, 0.02559, is a maximum turning point.
  • #1
JJV19
2
0
This is what i have so far

SA = 2(pi)r^2 + 2(pi)r(245.45 / (pi)r^2)

Derivative of SA = - 490.9r^2 + 4(pi)r

0 = -490.9r^2 + 4(pi)r

and I'm stuck at this step.. I've tried a bunch of things to try and solve for r but i can't seam to get a logical answer for the radius of the minimized pop can
 
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  • #2
minimise, that's something new, well to minimise why just not have a can at all? make r=0. this is one of the roots of the equation you gave me and it it the only min turnning point as d^2/dx^2 at the point is greater than 0. Moreso the other root 0.025 has d^2y/dx^2 less than 0 making it a maximum turnning point. The working follows:

0 = -490.9r^2 + 4(pi)r
0=r(-490.9r+4(pi))
hence r=0 or 490.9r=4(pi)
r=0 or r=0.02559
 
  • #3
JJV19 said:
This is what i have so far

SA = 2(pi)r^2 + 2(pi)r(245.45 / (pi)r^2)

Derivative of SA = - 490.9r^2 + 4(pi)r

0 = -490.9r^2 + 4(pi)r

and I'm stuck at this step.. I've tried a bunch of things to try and solve for r but i can't seam to get a logical answer for the radius of the minimized pop can

You didn't differentiate S correctly. According to your post,

[tex]S=2\pi r^{2} + 2\pi r(\frac{245.45}{\pi r^{2}}) = 2\pi r^{2} + \frac{490.9}{r}[/tex]

and

[tex]\frac{dS}{dr}=4\pi r - 490.9r^{2}[/tex]

Do you see where your error is?
 
  • #4
-490.9r^-2 not ^2
 
  • #5
JohnDuck said:
You didn't differentiate S correctly. According to your post,

[tex]S=2\pi r^{2} + 2\pi r(\frac{245.45}{\pi r^{2}}) = 2\pi r^{2} + \frac{490.9}{r}[/tex]

and

[tex]\frac{dS}{dr}=4\pi r - 490.9r^{2}[/tex]

Do you see where your error is?


thanks a lot both of you lol .. it's funny how the simple mistakes often cause the most grief
 

1. What does minimizing surface area of a can mean?

Minimizing surface area of a can refers to reducing the amount of space on the can's exterior that is exposed to the surrounding environment. This is typically done in order to decrease production costs or to make the can more efficient for storage or transportation.

2. Why is minimizing surface area important for cans?

Minimizing surface area is important for cans because it can reduce the amount of materials needed for production, leading to cost savings and a more environmentally friendly manufacturing process. It can also make the cans more compact and efficient for storage and transportation.

3. How is the surface area of a can calculated?

The surface area of a can is calculated by adding together the area of each individual face of the can. This includes the top and bottom circles, as well as the rectangular or cylindrical sides. The formula for finding the surface area of a cylinder is 2πr2 + 2πrh, where r is the radius and h is the height of the can.

4. What factors affect the surface area of a can?

The size and shape of the can, as well as the materials used, are the main factors that affect the surface area of a can. A larger can will have a larger surface area, and a can with a more complex shape will generally have a larger surface area as well. Additionally, the thickness of the material used to make the can will also impact its surface area.

5. How can the surface area of a can be minimized?

The surface area of a can can be minimized by using a more compact and efficient shape, such as a cylinder, and by reducing the thickness of the materials used to make the can. Additionally, using lightweight materials can also help to decrease the surface area of a can. Advanced manufacturing techniques, such as deep drawing, can also be used to create cans with minimal surface area.

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