Cylinder extreme values

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In summary, the conversation discusses a problem involving constructing a cylinder with the cheapest cost for the bottom and side parts. The conversation includes various mathematical calculations to determine the optimal radius and height of the cylinder. Ultimately, the conversation concludes with a request for assistance with the final calculations.
  • #1
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I've got kind of stuck with a problem that includes a cylinder.

First off, we know that the cylinder has a Volume of 100 litre ( V = 100 dm^3 )

but it get's abit of tricky since you are constructing this cylinder and you are trying to make the cheapest cylinder when the bottom parts of the cylinder costs 10€ / dm^2 and the side parts cost 5€ / dm^2.

HERE BEGINS MY "WORK"/GUESSING:

I've come to the conclusion that:

V = pi*r^2*h

so: pi*r^2*h = 100
also this makes: h = 100/ ( pi*r^2 )

so I figured I'd take:

10€ * pi*r^2 + 2*pi*r*h*5€

wich becomes:

10€ * pi*r^2 + 2*pi*r*100/ (pi*r^2) *5€

10€ * pi*r^2 + 2*100/r *5€

then take the derivate of that:

2*10€ * pi*r - r*1000/r^2

wich becomes:

2*10€ * pi*r - 1000/r

then you make then = 0 or just directly:

2*10€ * pi*r = 1000/r

then we try to make out what r is so we mix them around abit:

20 * pi * r = 1000/r

r^2 = 1000/(20*pi)

r = root(1000/(20*pi))

wich gives:

r = 3.989422...

wich it shouldn't, the real answer to r at this point should be something 2.52

and here is where I'm stuck, any help is appreciated :)
 
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  • #2
It should be [tex] \sqrt[3]{\frac{1000}{20\pi}} [/tex]
10€ * pi*r^2 + 2*100/r *5€

The derivative of this is [tex] 20\pi r -\frac{1000}{r^{2}} [/tex] not

[tex] 20\pi r - \frac{1000r}{r^{2}} [/tex]
 
  • #3
courtrigrad said:
It should be [tex] \sqrt[3]{\frac{1000}{20\pi}} [/tex]




The derivative of this is [tex] 20\pi r -\frac{1000}{r^{2}} [/tex] not

[tex] 20\pi r - \frac{1000r}{r^{2}} [/tex]

ohh... thanks a lot ^.^
 

What are extreme values of a cylinder?

The extreme values of a cylinder refer to the maximum and minimum values of its dimensions, such as height, radius, and volume.

How can extreme values be calculated for a cylinder?

The extreme values of a cylinder can be calculated by using the formula V=πr²h, where V is the volume, r is the radius, and h is the height.

Why are extreme values important in the study of cylinders?

Extreme values help us understand the range of possible dimensions for a cylinder and can be used to determine the maximum or minimum amount of material needed to construct it.

What factors can affect the extreme values of a cylinder?

The dimensions of a cylinder can be affected by various factors, such as the precision of measurements, the material used, and the manufacturing process.

How can extreme values impact the functionality of a cylinder?

The extreme values of a cylinder can impact its functionality by determining its capacity, strength, and stability. For example, a larger cylinder may have a higher capacity, but a thinner cylinder may be less stable.

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