Differential Calculus: Solving for Optimization

In summary, the students were trying to find an equation or function that would show the minimum cost for all volumes of any can. They eventually realized that v=minimum cost of v, and needed to find the optimal height and radius that gave the highest surface area to volume ratio or the smallest values for h and r.
  • #1
toasticles
13
0
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 realized 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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  • #2
The trick is, that now the volume can be anything, but it is still fixed.
At least, I suppose that the question they want you to answer is:
* For a can of fixed volume v, what are the optimal height and radius?

In that case, you can consider v as a constant, and express your answers in it, for example:
* the optimal dimensions are h = 12v and r = 6/h = 1/(2v)
 
  • #3
okay so i sort of see what you're saying there compu :)

however we need to find the minimum cost for any volume so yes the optimal height and radius that give the Highest surface area to volume ratio or the smallest values for h and r that give the Vi.e. (the smallest surface area that still gives the volume, v)

also why would v be fixed if it considered a variable if you graph it?

so if the minimum cost is meant to be found for any volume that means that cost and volume will both be dependant on height and radius so how do i write this as a function that will give me the graph where x-axis is the volume of a can and y-axis gives the corresponding optimal cost which is given by optimal height and radius?
 

1. What is differential calculus?

Differential calculus is a branch of mathematics that deals with the study of rates of change, or how a function changes over time. It involves finding the derivative of a function, which represents the instantaneous rate of change at a specific point.

2. What is optimization in differential calculus?

Optimization in differential calculus is the process of finding the maximum or minimum value of a function. This is done by finding the critical points of the function, which are the points where the derivative is equal to zero, and then determining whether these points correspond to a maximum or minimum value.

3. How do I solve for optimization using differential calculus?

To solve for optimization using differential calculus, you first find the derivative of the given function. Then, you set the derivative equal to zero and solve for the critical points. Finally, you use the second derivative test or the first derivative test to determine whether these critical points correspond to a maximum or minimum value.

4. What is the difference between a local and global maximum or minimum?

A local maximum or minimum is a point where the function reaches its highest or lowest value within a specific interval. A global maximum or minimum is the highest or lowest value of the entire function. In other words, a global maximum or minimum is also a local maximum or minimum, but the opposite is not always true.

5. What are some real-world applications of optimization using differential calculus?

Optimization using differential calculus has many real-world applications, such as maximizing profit in business, determining the most efficient route for delivery vehicles, and finding the best design for a product to minimize costs. It is also used in fields like physics, engineering, and economics to solve various optimization problems.

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