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1. Hey all, For my calculus class we were giving the problem of solving for the optimization of a tin can using differential calculus. The problem was to find the minimum cost for any tin can of any height(as well as using the equation for the tin we had). The surface area of the cylinder was considered to be 0.01cents/cm2 and the seams where only height was counted and the circumferences at the top and bottom of the can where 0.015cents/cm2.

2. So the cost function for the can i chose was C(r)=0.01(surface area)+0.015(Seams) which then amounts to C(r)=0.01((2Pir^2)+h(2Pir)+0.015(4Pir+h).

This is completely correct and what everyone got for the basic cost function. Now we differentiated this and let C=0 so we could find the minimum cost (using an 850ml volume can we see that:) if Volume=2pir^2*h then we are able to differentiate by substituting v with 850 and getting height in terms of radius hence one variable so:(expanded and simplified):

C'(r)=0.04Pir-(17r^-2)+0.06Pi-25.5Pir^-3

So we all found the minimum for a radius of 4.95 which is 5.797cents.

However the second part and insanely difficult to think about and figure out part was given to us moments later:

The challenge was to find an equation or function that shows the minimum cost FOR ALL volumes of any can So basically v=minimum cost of v

So on the x-axis one would have the volume and the y-axis one would have the minimum cost for that height.

None of us could get only 2 variables in the equation (v and r), where from the volume function you will always have volume substitued in and it just would'nt work

i hope this made sense and would reallllly like some help

3. Attempt at the optimized function for any can of any height finding the minimum cost:

minimum cost = (differential of cost = 0)

C'(r) = (dc/dr = 0)

here i realised i was going in the wrong direction cause h is no where in the function and i'm completely stuck none of the other maths B and C students can figure this out please help!!!

2. So the cost function for the can i chose was C(r)=0.01(surface area)+0.015(Seams) which then amounts to C(r)=0.01((2Pir^2)+h(2Pir)+0.015(4Pir+h).

This is completely correct and what everyone got for the basic cost function. Now we differentiated this and let C=0 so we could find the minimum cost (using an 850ml volume can we see that:) if Volume=2pir^2*h then we are able to differentiate by substituting v with 850 and getting height in terms of radius hence one variable so:(expanded and simplified):

C'(r)=0.04Pir-(17r^-2)+0.06Pi-25.5Pir^-3

So we all found the minimum for a radius of 4.95 which is 5.797cents.

However the second part and insanely difficult to think about and figure out part was given to us moments later:

The challenge was to find an equation or function that shows the minimum cost FOR ALL volumes of any can So basically v=minimum cost of v

So on the x-axis one would have the volume and the y-axis one would have the minimum cost for that height.

None of us could get only 2 variables in the equation (v and r), where from the volume function you will always have volume substitued in and it just would'nt work

i hope this made sense and would reallllly like some help

3. Attempt at the optimized function for any can of any height finding the minimum cost:

minimum cost = (differential of cost = 0)

C'(r) = (dc/dr = 0)

here i realised i was going in the wrong direction cause h is no where in the function and i'm completely stuck none of the other maths B and C students can figure this out please help!!!

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