Optimizing Cost for Given Volume of a Cylinder

In summary, to find the optimal radius and height for any volume with minimum cost, the following steps can be taken: first, replace "v" with \pi r^2 h in the cost equation. Then, differentiate the cost equation with respect to both r and h, treating h as a constant. Set both derivatives equal to 0 and solve for r and h. Finally, use the optimal values for r and h to find the minimum cost for the given volume.
  • #1
toasticles
13
0
if Cost(r)=0.01(2[tex]\pi[/tex]r2+v/[tex]\pi[/tex]r2(2[tex]\pi[/tex]r))+0.015(4[tex]\pi[/tex]r+v/[tex]\pi[/tex]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 ^^
 
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  • #2
The first thing I would do is go ahead and replace "v" by [itex]\pi r^2 h[/itex]:
[itex]cost(r, h)= 0.01(2\pi r^2+[v/(\pi r^2)][/itex][itex](2\pi r)+0.015(4\pi r+v/(\pi r^2)[/itex]
[itex]cost(r, h)= 0.01(2\pi r^2+[(\pi r^2h)/(\pi r^2)][/itex][itex](2\pi r)+0.015(4\pi r+(\pi r^2 h)/(\pi r^2)[/itex]
[itex]cost(r, h)= 0.01(2\pi r^2+ h(2\pi r)+ 0.015(4\pi r+ h))[/itex]

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

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
HallsofIvy said:
[itex]cost(r, h)= 0.01(2\pi r^2+ h(2\pi r)+ 0.015(4\pi r+ h))[/itex]

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 separated completely by the addition sign? (i'm only first year calculus) so if it's a law or something please explain :D
 
  • #4
toasticles said:
if Cost(r)=0.01(2[tex]\pi[/tex]r2+v/[tex]\pi[/tex]r2(2[tex]\pi[/tex]r)+0.015(4[tex]\pi[/tex]r+v/[tex]\pi[/tex]r2)
v=hpir2

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

1. What is a multivariable derivative?

A multivariable derivative is a mathematical concept that describes how a function changes with respect to multiple variables. It is similar to a regular derivative, which measures the rate of change of a function with respect to a single variable, but takes into account the effect of multiple variables on the function.

2. Why are multivariable derivatives important?

Multivariable derivatives are important because they allow us to analyze the behavior of functions with multiple variables. They are used in various fields of science and engineering, such as physics, economics, and engineering, to model and understand complex systems.

3. How do you calculate a multivariable derivative?

To calculate a multivariable derivative, we use partial derivatives, which measure the rate of change of a function with respect to each individual variable while holding all other variables constant. We then combine these partial derivatives using a specific formula, depending on the type of multivariable derivative being calculated.

4. What is the difference between a partial derivative and a multivariable derivative?

A partial derivative measures the rate of change of a function with respect to a single variable, while holding all other variables constant. A multivariable derivative, on the other hand, takes into account the effect of multiple variables on the function. It is a more comprehensive and complex concept that is used to analyze functions with multiple variables.

5. In what real-world situations are multivariable derivatives used?

Multivariable derivatives are used in a wide range of real-world situations, such as optimizing production processes in manufacturing, modeling stock market trends in finance, and predicting weather patterns in meteorology. They are also used in physics to analyze the motion of objects in three-dimensional space and in economics to study the relationship between multiple variables, such as supply and demand.

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