Max Volume Parcel w/ Triangular Prism: Canada Post

  • Thread starter Thread starter dimpledur
  • Start date Start date
  • Tags Tags
    Volume
dimpledur
Messages
193
Reaction score
0

Homework Statement


Canada Post accepts international parcels whose (Length+Girth) is less than or equal to 2 meters, and Length is less than or equal to 1 meter. Girth is defined as the cross section. We wish to ship a parcel of the shape of a triangular prism of length l meters. The cross section is a right triangle with catheti of lengths a and b meters. Assume the package walls are thin. What is the maximal volume of a parcel?


Homework Equations


Let
leg_1=a=y, leg_2=b=z, length=x
I provided a drawing via paint for you to envision my take on the problem
Untitled-7.png

Hence,
V(x, y, z)=\frac{1}{2}xyz
Boundaries:
(x+y+z+\sqrt{x^2+y^2})\leq{2}, x\leq{1}

The Attempt at a Solution


\nabla{f}(x, y, z)=\frac{1}{2}(yzi+xzj+xyk)
and hence there exists a critical point at (0, 0, 0).

Next, I get partially lost. Should I be finding second partial derivatives of the boundary and then evaluating the Hessian matrix to determine extremes on the boundary?
 
Physics news on Phys.org
dimpledur said:

Homework Statement


Canada Post accepts international parcels whose (Length+Girth) is less than or equal to 2 meters, and Length is less than or equal to 1 meter. Girth is defined as the cross section. We wish to ship a parcel of the shape of a triangular prism of length l meters. The cross section is a right triangle with catheti of lengths a and b meters. Assume the package walls are thin. What is the maximal volume of a parcel?


Homework Equations


Let
leg_1=a=y, leg_2=b=z, length=x
I provided a drawing via paint for you to envision my take on the problem
Untitled-7.png

Hence,
V(x, y, z)=\frac{1}{2}xyz
Boundaries:
(x+y+z+\sqrt{x^2+y^2})\leq{2}, x\leq{1}

The Attempt at a Solution


\nabla{f}(x, y, z)=\frac{1}{2}(yzi+xzj+xyk)
and hence there exists a critical point at (0, 0, 0).

Next, I get partially lost. Should I be finding second partial derivatives of the boundary and then evaluating the Hessian matrix to determine extremes on the boundary?

Critical points of V are irrelevant in this problem, due to the presence of constraints. Here is a simpler example: what is the maximum of f(x) =x^2, subject to 0 <= x <= 1? Obviously, the max is at x = 1, but the derivative of f is 2 at that point, not zero.

RGV
 
hmm, and when I tried finding critical points within the bounded portion the gradient told me there were not. It doesn't seem logical to just start plugging numbers in, as there are three variables that depend on girth.
 
dimpledur said:
hmm, and when I tried finding critical points within the bounded portion the gradient told me there were not. It doesn't seem logical to just start plugging numbers in, as there are three variables that depend on girth.

You have forgotten three very important restrictions: x >= 0, y >= 0, z >= 0. If you were to submit your problem to a computer package, but without these restrictions, you would get a nonsensical solution: you can let x and y go to -infinity, and fix z = 1. The both of your written restrictions are satisfied, and V --> + infinity (which is certainly as much of a maximum as you could want!).

Anyway, your complete problem is:
max V(x,y,z)
subject to g(x,y,z) <= 2, x <= 1 and x, y, z >= 0.
(Here, V and g are the functions you wrote.)

You can solve this problem using the Karush-Kuhn-Tucker conditions (essentially, the Lagrangian conditions extended to inequality constraints). If all you want is a numerical solution, and don't care how you get it, you can use the EXCEL Solver tool to get a solution in an EXCEL spreadsheet. Other spreadsheets (including free, open source ones) also have solver tools.

Google 'Karush-Kuhn-Tucker' and 'Excel Solver' for more information.

RGV
 
Okay, I have had a bit more experience with Lagrange multipliers, hence here is my next attempt.
The maximum must occur on the boundary of,
(x+y+z+\sqrt{z^2+y^2})={2}
Hence, let
L=0.5xyz+\lambda{(}x+y+z+\sqrt{y^2+z^2-2})
The critical points of L are determined via
L_1=0.5yz+\lambda
L_2=0.5xz+\lambda{+}\frac{2y}{\sqrt{y^2+z^2}}
L_3=0.5xy+\lambda{+}\frac{2z}{\sqrt{y^2+z^2}}
L_4=x+y+z+\sqrt{x^2+y^2}-2
Is there any easy way to compute the solutions without using computational software?
 
I guess the next step would be to solve the system of equations for the four unknowns, since there are four equations.

ALso was informed that there was an easier method to solving this question.
 
Last edited:
There are two things I don't understand about this problem. First, when finding the nth root of a number, there should in theory be n solutions. However, the formula produces n+1 roots. Here is how. The first root is simply ##\left(r\right)^{\left(\frac{1}{n}\right)}##. Then you multiply this first root by n additional expressions given by the formula, as you go through k=0,1,...n-1. So you end up with n+1 roots, which cannot be correct. Let me illustrate what I mean. For this...
Back
Top