# Thread: MAX volume given array of constraints

1. ## MAX volume given array of constraints

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 perimeter of 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?

$leg_1=a=y, leg_2=b=z, length=x$
I provided a drawing via paint for you to envision my take on the problem

Hence,
$V(x, y, z)=\frac{1}{2}xyz$
Boundaries:
$(x+y+z+\sqrt{y^2+y^z})\leq{2}\\ x\leq{1}\\ x\geq{0}\\ y\geq{0}\\ z\geq{0}$

The critical points of the gradient of the volume equation comes up with (0, 0, 0), hence it is of no use. Okay, I have not had much experience with Lagrange multipliers, but here is my attempt.

$(x+y+z+\sqrt{z^2+y^2})={2}$
Hence, let
$L=0.5xyz+\lambda{(}x+y+z+\sqrt{y^2+z^2}-2)$
The critical points of L are determined via
$L_1=0.5yz+\lambda$
$L_2=0.5xz+\lambda{+}\frac{2y}{\sqrt{y^2+z^2}}$
$L_3=0.5xy+\lambda{+}\frac{2z}{\sqrt{y^2+z^2}}$
$L_4=x+y+z+\sqrt{x^2+y^2}-2$

At this point the system of equations look quite nasty, and moreoever I have been told that there is an easier route to solving this problem. Can someone suggest something?

2. ## Re: MAX volume given array of constraints

I am a bit confused why the "cross section" is the perimeter of the triangle. Cross section tends to refer to area. If I were doing the problem, my understanding of the constraints would have been:

$x+\frac{1}{2}yz \le 2$

If this is the case, then:

$L=\frac{1}{2}xyz + \lambda(x+\frac{1}{2}yz-2)$
$L_x = \frac{1}{2}yz + \lambda$
$L_y = \frac{1}{2}xz + \frac{1}{2}z\lambda$
$L_z = \frac{1}{2}xy + \frac{1}{2}y\lambda$
$L_\lambda = x+\frac{1}{2}yz-2$
Now the system of equations looks much easier. Are you sure you really wanted the girth to be the perimeter?

3. ## Re: MAX volume given array of constraints

Question is quoted as is. " Canada post accepts international parcels whose (Length+Girth) $\leq$2."

4. ## Re: MAX volume given array of constraints

Originally Posted by quantoembryo
Question is quoted as is. " Canada post accepts international parcels whose (Length+Girth) $\leq$2."
Agreed. Now, girth = "cross section". You wrote that the "cross section" is the perimeter of the triangle with catheti $a,b$. I am suggesting that the "cross section" may be the area of that triangle instead. See Cavalieri's principle, Fubini's principle, et. al. for other instances where the term "cross section" refers to the "cross sectional area", not the "cross sectional perimeter". So, if you know that the girth is supposed to be the perimeter, I will think about the problem more. Otherwise, it works much more easily with area.

5. ## Re: MAX volume given array of constraints

Okay, thanks for the response. I had never dealt with "girth" so I googled it and it stated that the girth is the perimeter.

6. ## Re: MAX volume given array of constraints

Ok, that makes sense! I did not google girth. I simply used what I knew about cross sections in general math. Ok, perimeter it is!

I will need to think about the problem some more.

7. ## Re: MAX volume given array of constraints

Just realized I left out an important portion of the original problem. It states "girth is defined as the perimeter of the cross section". Sorry about that! Not sure how I could have missed that.. I have the flu so I am not operating on all cylinders.

8. ## Re: MAX volume given array of constraints

Let $z=ky$ where $k>0$. Now, you have:

$V(x,y,k) = \frac{1}{2}kxy^2$
$x+y+ky+\sqrt{y^2+k^2y^2} \le 2$

So
$x+(k+1)y + y\sqrt{k^2+1} \le 2$

Now, assume that the $\le$ is an equals, and you can solve for $x$ and get an equation for volume in $y,k$

9. ## Re: MAX volume given array of constraints

$V(x,y,k) = \frac{1}{2}kxy^2$
$x=-y-ky-\sqrt{y^2+k^2y^2}+2$

Hence,
$V(x,y,k) = \frac{1}{2}k(-y-ky-\sqrt{y^2+k^2y^2}+2)y^2$

$V(x,y,k) =-\frac{1}{2}ky^3-\frac{1}{2}k^2y^3-\frac{1}{2}k^2y^3-\frac{1}{2}ky^2\sqrt{y^2+k^2y^2}+ky^2$

From here you're suggesting that this is simpler to evaluate?

10. ## Re: MAX volume given array of constraints

I believe it is.

11. ## Re: MAX volume given array of constraints

$V(x,y,k) =-\frac{1}{2}ky^3-k^2y^3-\frac{1}{2}ky^2\sqrt{y^2+k^2y^2}+ky^2$

$V_1=0$

$V_2=-\frac{3}{2}ky^2-3k^2y^2-ky\sqrt{y^2+k^2y^2}-\frac{ky^3+k^3y^3}{2\sqrt{y^2+k^2y^2}}+2ky$

$V_3=-\frac{1}{2}y^3-2ky^3-\frac{1}{2}y^2\sqrt{y^2+k^2y^2}-\frac{k^2y^3}{\sqrt{y^2+k^2y^2}}$

12. ## Re: MAX volume given array of constraints

Sorry, I didn't give you one last step to simplify...

The volume is also $xA$ where $A$ is the area of the triangle. Because that area does not depend on $x$, you can maximize that area first in order to solve for $k$. I meant to do that before taking a derivative. You wind up with $k=1$.

13. ## Re: MAX volume given array of constraints

So first I maximize the area of the triangle and hence solve for k, then I can plug k into the partials I have already solved and thus determine y?

14. ## Re: MAX volume given array of constraints

No, those partials are going to make it wayyyy more difficult than you need. Plug it into the original volume equation. You wind up with an equation for volume based entirely on $y$.

15. ## Re: MAX volume given array of constraints

Originally Posted by quantoembryo
$V(x,y,k) = \frac{1}{2}kxy^2$
$x=-y-ky-\sqrt{y^2+k^2y^2}+2$

Hence,
$V(x,y,k) = \frac{1}{2}k(-y-ky-\sqrt{y^2+k^2y^2}+2)y^2$

$V(x,y,k) =-\frac{1}{2}ky^3-\frac{1}{2}k^2y^3-\frac{1}{2}k^2y^3-\frac{1}{2}ky^2\sqrt{y^2+k^2y^2}+ky^2$

From here you're suggesting that this is simpler to evaluate?

Hence,
$V(y) = \frac{1}{2}(-2y-y\sqrt{2}+2)y^2$

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