Calculus Optimization problems

In summary, the conversation was about solving an optimization problem for finding the cheapest rectangular storage container with an open top. The volume, length, and width of the container were given and the cost of materials for the base and sides were also provided. The conversation included discussions on writing equations, using substitution, finding derivatives, and determining critical values to find the minimum cost. Ultimately, the conversation ended with the conclusion that the minimum cost can be achieved at w=cube root (9/2).
  • #1
scorpa
367
1
Hello Everyone,

I'm working on optimization problems right now, and let me tell ya, if you thought I was horrible at everything else wait until you see me attempt one of these :grumpy: I have three questions but I'm hoping if I just get help with the one I will be able to get the processes for the rest on my own. Here it goes:

A rectangular storage container with an open top is to have a volume of 10m^3. The length of the base is twice that of the width. Material for the base costs $10 per square meter and and $6 per square meter for the side. Find the cost of materials for the cheapest such container :yuck:

OK So first off all I wrote down all my givens and drew myself a little picture which I had hoped would help me visualize the problem. Now I know that the volume of the container is 10 and that V=LWH, but that the length is twice the width allowing me to get rid of one of the variables. Leaving me with 10=2w^2 *h. Ic ould then get rid of the h by saying that h=(5/w^2).

P=2l +2w +h
P = 2(2w) +2w +5/w^2

At this point I end up getting weird numbers, I have a feeling that part of the problem is that I never factored in that the box has an open top but I'm not sure how to go about doing that. I've tried to show what I've done and my reasons for doing it, but unfortunately I know its wrong. Any help you can give I would really appreciate, thanks a lot!
 
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  • #2
Your price equation is a bit odd, would you agree that it should be:

Total price = (area of base)x$10 + 2x(area of long side)x$6 + 2x(area of short side)x$6.

Your equation involving the relationship between the side dimensions look good.
 
  • #3
OK, I hadn't actually gotten to the price part yet, that was actually for perimeter but maybe that wasn't actually needed? I will keep trying to get it just using the price equation you suggested.Thanks
 
  • #4
OK, no I guess I'm still lost. I can't seem to fit anything in. I'm hopelessly confused.
 
  • #5
In an optimization problem, you first need to write the equation for the parameter you want to optimize (obviously), then differentiate it to find its max or min. There must be other constraints involved as well. Here, the constraints are (1) volume must be 10 (2) length must be twice the width. You want to optimize price as a function of 3 basic paramters (L, W, H).

You have a relationship between two. Try some substitution to see if you can get the price in terms of only 1 variable, then differentiate.
 
  • #6
OK thanks ill keep working on it!
 
  • #7
OK so, I went ahead with that price formula and stuck in the numbers as follows:

(2w^2)(10)+(2)(2wh)(6)+(2)(2h)(6), i then subsituted in my h value of 5/w^2

Then I simplifed and took the derivative so I could figure out the max and mins. My derivative was (w^-2)(40w^3-180).

Does this sound right so far? This way I ended up with roots of 0 and cube root 4.5 for my critical values which just doesn't seem right. I'm starting to wonder if the fact that the box has an open top is displayed incorrectly in my work?
 
  • #8
From what I get, the price equation simplifies to:

[tex] P = 44W^2 + \frac{60}{W}[/tex]
 
  • #9
Oh ok, I guess I'll keep trying to see if I can get that to. Thanks, I'll keep at it
 
  • #10
Ok I never got the price equation to simplify to what you did, but I did finish the question. Maybe if i post what I did you will be able to see where I went wrong.

Ok my equation for volume, and my way of isolating h hasn't changed from above. And according to the price equation you listed I end up getting the following for price:

P = 20w^2+24wh+12wh
P=20w^2 +36wh

Then to get rid of the h I substitued in h=5/w^2 which I determined above which gets me:

P = 20w^2 +36w(5/w^2)
P = 20w^2 + 180W^-1

Then I found the derivative of this so I could find the minimum cost.

P' = 40w-180W^-2
P' = 20w^-2(2w^3-9)

Then I found that the critical values of the equation are w=0 and w=cube root (9/2)

Do I have it now?
 
  • #11
My apologies for my calculations earlier. Apparently, I cannot get the same result I did back then ( I kind of did it quickly in my head which failed me miserably ).

Your answer seems very right! I think you got it ^_^
 
  • #12
Oh thank you so much! You have no idea how much I appreciate your help! Now all I have to do is go back and figure out what the actual cost is going to be. Thanks a lot!
 

1. What is a calculus optimization problem?

A calculus optimization problem involves using calculus techniques to find the maximum or minimum value of a function. This can be used to solve real-world problems such as maximizing profits or minimizing costs.

2. What is the first step in solving a calculus optimization problem?

The first step is to define the variables and constraints of the problem. This includes identifying the function to be optimized, any given constraints, and the domain of the problem.

3. How do you find the critical points in a calculus optimization problem?

The critical points can be found by taking the derivative of the given function and setting it equal to zero. This will give you the x-values that correspond to the maximum or minimum points of the function.

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

A local maximum is the highest point on a particular interval, while a global maximum is the highest point on the entire function. A local maximum may not necessarily be the overall highest point of the function.

5. How do you determine if a critical point is a maximum or minimum?

To determine if a critical point is a maximum or minimum, you can use the second derivative test. If the second derivative is positive at the critical point, then it is a minimum. If the second derivative is negative, then it is a maximum.

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