Multivariable Derivates?

1. Aug 23, 2010

toasticles

if Cost(r)=0.01(2$$\pi$$r2+v/$$\pi$$r2(2$$\pi$$r))+0.015(4$$\pi$$r+v/$$\pi$$r2)
v=hpir2
how does one show for the optimal radius/height for any volume i.e. where the cost will be the minimum for that volume?

all my attempts have ended with 0=0

help please? i'll give you cookehz ^^

Last edited: Aug 23, 2010
2. Aug 23, 2010

HallsofIvy

The first thing I would do is go ahead and replace "v" by $\pi r^2 h$:
$cost(r, h)= 0.01(2\pi r^2+[v/(\pi r^2)]$$(2\pi r)+0.015(4\pi r+v/(\pi r^2)$
$cost(r, h)= 0.01(2\pi r^2+[(\pi r^2h)/(\pi r^2)]$$(2\pi r)+0.015(4\pi r+(\pi r^2 h)/(\pi r^2)$
$cost(r, h)= 0.01(2\pi r^2+ h(2\pi r)+ 0.015(4\pi r+ h))$

Now differentiate with respect to both r and h:
Treating h as if it were a constant,
$$cost_r(r, h)= 0.01(4\pi r+ 2\pi h+ 0.015(4\pi))$$
$$cost_h(r, h)= 0.01(2\pi r+ 0.015)$$

To find the optimum height, set those both equal to 0 and solve the two equations for r and h. (Since there is no "h" in the second equation , it is particularly easy to solve it for r. Although it looks to me like there is a sign error- that will give a negative value for r.)

3. Aug 23, 2010

toasticles

i don't think this is right or if it is how did you get it so that the second part(0.15(4pir+h) is also multiplied by 0.01 as the 0.01 and 0.015 "segments" of the equation are seperated completely by the addition sign? (i'm only first year calculus) so if it's a law or something please explain :D

4. Aug 23, 2010

lanedance

i think its because your original brackets were a little ambiguous - why not use the method but with what you've got?