Finding the Cheapest Cylinder with Extreme Values: Help Needed!

[me_maths]

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 :)

 

[courtrigrad]

It should be $$\sqrt[3]{\frac{1000}{20\pi}}$$

10€ * pi*r^2 + 2*100/r *5€

The derivative of this is $$20\pi r -\frac{1000}{r^{2}}$$ not

$$20\pi r - \frac{1000r}{r^{2}}$$

 

[me_maths]

It should be $$\sqrt[3]{\frac{1000}{20\pi}}$$




The derivative of this is $$20\pi r -\frac{1000}{r^{2}}$$ not

$$20\pi r - \frac{1000r}{r^{2}}$$

ohh... thanks a lot ^.^