Optimization, Minima, Open Top Box

In summary: RSIn summary, to minimize the amount of material used in a box with a square base and no top, with a volume of 10000 cm^3 and a smallest dimension of 5 cm, the dimensions of the box should be x = 27.144 cm and y = 6.786 cm. Similarly, for a cylindrical can to hold 500 cm^3 of apple juice with a height between 6 and 15 cm, the minimum amount of material used can be achieved by setting the derivative of the surface area equation to zero and solving for r, resulting in a radius of approximately 3.479 cm.
  • #1
rabbitstorm
6
0
Optimization, Minima, new question:Sheet Alluminum

Homework Statement


A box with a square base and no top must haave a volume of 10000 cm^3. If the smallest dimension in any direction is 5 cm, then determine the dimensions of the box that minimize the amount of material used.


Homework Equations


Volume: x^2y
Surface Area: x^2+4xy



The Attempt at a Solution


Let x represent the length and width of the box
Let y represent height

x>5

I drew a neat little drawing of the box and labeled it according to the statements above.

V=x^2y
10000=x^2y
y=10000/x^2


SA=x^2+4xy
=x^2+4x(10000/x^2)

Now I think that I need to set the SA equation to zero then differentiate but I can't quite remember what I'm doing with it. I'd appreciate anyhelp that could be offered.


~Thanks!
 
Last edited:
Physics news on Phys.org
  • #2
Don't forget to subtract the area of the top from the surface area equation.
 
  • #3
He never added it lol.

Your on the right track. [tex]SA=x^2+\frac{40000}{x}[/tex]

Differentiate with respect to x and set equal to zero. you get [tex]2x=\frac{40000}{x^2}[/tex] Take the x^2 over and we get 2x^3=40000, x^3=20000. Take the cube root, we get around 27.144cm. Sub that value Into the 10000=x^2 y to get y.
 
  • #4
wow.. Thanks that make sense I think, been a couple months since we did this in class.

Thanks
~RS

So then x=27.144 cm and y=6.786 cm.
Don't I also sub 5 from x>5 into the equation to tie up loose ends and make sure that it really is 27 cm? oops wrong equation y=13 and change.
 
Last edited:
  • #5
A cylindrical can is to hold 500 cm^3 of apple juice the design must take into account that the height must be between 6 and 15 cm, inclusive. How should the can be constructed so that a minimum amount of material will be used in the construction? assume no waste.

V is volume and SA is surface area

so far this is what I have:

15>h>6

v=(pi)r^2h
500=(pi)r^2h
500/(pi)r^2=h

SA=2(pi)r(500/(pi)r^2) + 2(pi)r^2

Now I believe I need to find the derivative and set it to zero then solve for r.

The problem is that I can't quite figure out how I'm supposed to find the derivative if someone could give me a hand understanding the process as I have forgotten and my teacher is busy with the latest stuff.
 
  • #6
No, your requirement is that x and y must both be greater than 5 which is true.
 
  • #7
HallsofIvy said:
No, your requirement is that x and y must both be greater than 5 which is true.


Sorry I forgot to say that I finally got that question finished and it seems to be correct, I had been struggling with it for the last few days and my teacher wasn't able to explain it in a way that helped. Thanks to Gib Z it clicked and I figured it out as far as boxes go, but when I see derivatives and pi I get thrown off.

Thanks,
 

1. What is optimization?

Optimization is the process of finding the best possible solution for a given problem. It involves maximizing or minimizing a particular objective function while satisfying any constraints that may be present.

2. What is a minimum?

A minimum is the lowest point or value of a function. It is the optimal solution in the case of a minimization problem, where the goal is to find the lowest possible value of the objective function.

3. How is optimization used in science?

Optimization has many applications in science, including engineering, economics, physics, and biology. It is used to find the optimal design of structures and systems, determine the most efficient use of resources, and understand natural phenomena.

4. What is an open top box?

An open top box is a type of container that has an opening at the top, allowing for easy access to the contents inside. It is often used in packaging and storage, as well as in mathematical models to represent a constrained optimization problem.

5. How is optimization related to an open top box?

In the context of mathematics, optimization can be used to find the optimal dimensions of an open top box that will maximize its volume while using a fixed amount of material. This is an example of a constrained optimization problem, where the constraints are the fixed amount of material and the desired volume of the box.

Similar threads

  • Calculus and Beyond Homework Help
Replies
11
Views
3K
Replies
7
Views
1K
  • Calculus and Beyond Homework Help
Replies
7
Views
1K
  • Calculus and Beyond Homework Help
Replies
3
Views
2K
  • Calculus and Beyond Homework Help
Replies
4
Views
1K
  • Calculus and Beyond Homework Help
Replies
2
Views
2K
  • Calculus and Beyond Homework Help
Replies
1
Views
1K
  • Calculus and Beyond Homework Help
Replies
3
Views
1K
  • Calculus and Beyond Homework Help
Replies
17
Views
3K
  • Calculus and Beyond Homework Help
Replies
5
Views
4K
Back
Top