can someone help me out with this calculus 1 problem?

[skim milk]

it's a related rates problem

A soda can is to hold 12 fluid ounces. Find the dimensions (radius and height) that will minimize the amount of material used in its construction, assuming that the thickness of the material is uniform.

thanks....

 

[TuxDave]

Originally posted by: fritolays
it's a related rates problem

A soda can is to hold 12 fluid ounces. Find the dimensions (radius and height) that will minimize the amount of material used in its construction, assuming that the thickness of the material is uniform.

thanks....

Well, you're given two equations:
pi*(radius^2)*height = 12 : restriction due to the volume of the soda can
amount of material = pi*(radius^2) + 2*pi*radius*height : number you must minimize

Using the first equation, you can replace height in terms of radius. That way you have the amount of material a function of only one variable. If you take the derivative, you should be able to minimize it.

edit: I actually have no idea what 12 fluid ounces is. Is it equal to 12 cubic centimeters?

 

[skim milk]

Originally posted by: TuxDave

Originally posted by: fritolays
it's a related rates problem

A soda can is to hold 12 fluid ounces. Find the dimensions (radius and height) that will minimize the amount of material used in its construction, assuming that the thickness of the material is uniform.

thanks....

Well, you're given two equations:
pi*(radius^2)*height = 12 : restriction due to the volume of the soda can
amount of material = pi*(radius^2) + 2*pi*radius*height : number you must minimize

Using the first equation, you can replace height in terms of radius. That way you have the amount of material a function of only one variable. If you take the derivative, you should be able to minimize it.

so you're saying.... solve the first equation in terms of radius
plug it in into the second equation for h..... take the first derivative?

 

[TuxDave]

Yup, take the first derivative and set it equal to zero. Of course if you want to be 100% accurate, if you find the derivative and set it to zero, that'll find a max or min. You'd probably have to take the derivative AGAIN to prove that it's a minimum. (2nd Derivative must be positive at the value that you want). That's the first solution that I see. Dunno why it's called related rates though.... *shrug*

 

[skim milk]

Originally posted by: TuxDave
Yup, take the first derivative and set it equal to zero. Of course if you want to be 100% accurate, if you find the derivative and set it to zero, that'll find a max or min. You'd probably have to take the derivative AGAIN to prove that it's a minimum. (2nd Derivative must be positive at the value that you want). That's the first solution that I see. Dunno why it's called related rates though.... *shrug*

ok thanks, i will try it

btw, i have a question about second derivative

let's say on a graph, the second derivative of a function is < 0 if -2 < x < 2 and the second derivative of a function is > 0 if x<-2 or x>-2
what is this exactly saying about the graph?

thanks....

 

[Paul180]

When you solve the derivative of the function when it = 0, you are finding the maximum or minimum points. You only have to worry about what the graph looks like if you get more than one answer for the solution to the derivative = 0, because then you might have a maximum or you might have a minimum.

The first derivative is the slope of the function, and the second derivative is the change in the slope... So when the second derivative is 0, its a point of inflection, when its > 0 then its concave up ( a minimum) and when its < 0 its concave down ( a maximum). The second derivative test is merely a way of proving that at that critical point there exists a maximum or minimum. You should not have to worry about this because it really has nothing to do with your problem... you would only need this if you had to prove your answer. This subject matter can be quite confusing...

Hope it makes sense and helps...

 

[Vadatajs]

<grabs a beer> 12oz=355ml=355cc