What Are the Dimensions of the Least Expensive Conical Frustum Container?

[Morfe]

Hi there!
Kindly help me to solve the problem below.

A company is using frustum of a cone containers for their products. What are the dimensions of the least expensive container that can hold 300 cubic cm? Use Lagrange Multipliers to solve the problem.

Thanks.

 

[BvU]

Well, what do you know already to set up this exercise ?

 

[Gaussian97]

Hi there!
Kindly help me to solve the problem below.

A company is using frustum of a cone containers for their products. What are the dimensions of the least expensive container that can hold 300 cubic cm? Use Lagrange Multipliers to solve the problem.

Thanks.

Well, to do Lagrange multipliers you need two things, first of all, some functions that you want to minimize, in your case is only one function, the cost, so find a function that gives you the cost of any frustum of a cone.
Then you also need a set of function to constrain, in your case again is only one, that is the volume, so you also need to find a function that gives you the volume of any frustum.

Try to find these two functions first before continue.

 

[Morfe]

Well, to do Lagrange multipliers you need two things, first of all, some functions that you want to minimize, in your case is only one function, the cost, so find a function that gives you the cost of any frustum of a cone.
Then you also need a set of function to constrain, in your case again is only one, that is the volume, so you also need to find a function that gives you the volume of any frustum.

Try to find these two functions first before continue.

Hi I already generated the system of equations from the partial derivatives (please see attachment). My problem now is on how to compute the system of equation with complex equations. Thanks. Hope you can help me with this.

 

[BvU]

I see three unknowns ($H, h, k$) (*). So I expect three equations. Are there pages missing ?

(*) Well, you make things difficult by bringing $L$ into the expressions. But it's not a new unknown.

 

[Morfe]

I see three unknowns ($H, h, k$). So I expect three equations. Are there pages missing ?

There are four variables (H,h,k,λ) and I was able to generate four equations. The four equations are on the attachment. Thanks.

 

[BvU]

Oops, more pages! My bad.
So what's the problem ?

 

[Morfe]

Oops, more pages! My bad.
So what's the problem ?

My problem now is on how to compute the system of equation with complex equations.
Thanks for the help.

 

[Morfe]

My problem now is on how to compute the system of equation with complex equations.
Thanks for the help.

Is it possible to solve the system of equation found on the attachment? Or are there any software that can solve it?
Thanks.

 

[Gaussian97]

Try to add the equation (1) and (2)

 

[zinq]

The problem is not stated clearly enough to be solved.

Exactly which part(s) of the frustum of a cone need to be manufactured? Is it just the portion of the cone per se? The cone and the larger disk? The cone and the smaller disk? The cone and both disks?

And presumably the cost is proportional to the total surface area.

These things need to clarified before charging ahead blindly to solve the problem, since it's always useful to know which problem you want to be solving.

 

[Morfe]

The problem is not stated clearly enough to be solved.

Exactly which part(s) of the frustum of a cone need to be manufactured? Is it just the portion of the cone per se? The cone and the larger disk? The cone and the smaller disk? The cone and both disks?

And presumably the cost is proportional to the total surface area.

These things need to clarified before charging ahead blindly to solve the problem, since it's always useful to know which problem you want to be solving.

Hi!
The portions to be considered in the problem are the lateral surface of the frustum, and the upper and lower bases.
Thanks.

 

[Fred Wright]

Your problem is minimizing the surface area of a conical frustum w.r.t a given volume. The surface area and volume of the frustum are$$
A=\pi (r_b + r_f)\sqrt{h^2+(r_b - r_f)^2}\\
V=\frac{\pi h}{3}(r_b^2+r_br_f+r_f^2)$$where $r_b$ is the radius of the base, $r_f$ is the radius of the frustum and $h$ is the height. If you use the Lagrangian multiplier method your Lagrangian is$$
\mathcal {L}(r_b,r_f,h,\lambda) = \pi (r_b + r_f)\sqrt{h^2+(r_b - r_f)^2} +\lambda (\frac{\pi h}{3}(r_b^2+r_br_f+r_f^2)-\gamma)$$ where $\gamma$ is the constraining volume and $\lambda$ is the Lagrangian multiplier. We get four equations in four unknowns:$$
\frac{\partial {\mathcal {L}(r_b,r_f,h,\lambda)}}{\partial {r_b}}=0\\
\frac{\partial {\mathcal {L}(r_b,r_f,h,\lambda)}}{\partial {r_f}}=0\\
\frac{\partial {\mathcal {L}(r_b,r_f,h,\lambda)}}{\partial {h}}=0\\
\frac{\partial {\mathcal {L}(r_b,r_f,h,\lambda)}}{\partial {\lambda}}=0$$
On performing these derivatives and eliminating variables one is quickly confronted with an intractable problem in algebra. I therefore propose that minimizing the perimeter of the frustum cross section w.r.t. area is, by rotational symmetry, equivalent to minimizing the surface area w.r.t. volume. The minimal perimeter of a rectangle is a square and the minimal perimeter of a right triangle is a $\frac{\pi}{4}$ equilateral triangle. Let $x$ be the side of a square. We find:$$
r_b=\frac{3}{2}x\\
r_f=\frac{1}{2}x\\
h=x$$and solving for x w.r.t to volume$$
\gamma = \frac{\pi}{3}x^2(\frac{9}{4}+\frac{3}{4} +\frac{1}{4})\\
x=\sqrt{\frac{12\gamma}{13\pi}}$$Using the Lagrangian multiplier method to solve this problem is like using a bulldozer to do the job of a garden trowel.