Box optimization problem

J

jdb

New member
Joined
Oct 9, 2010
Messages
2
A simple way to model the construction of an oil tanker is to start with a large rectangle sheet of steel that is x feet wide and 3x feet long. Now cut a smaller square that is y feet on a side out of each corner of the larger sheet and fold up and weld the sides of the steel sheet to make a traylike structure with no top.

I've shown that the volume can be represented by the equation:

V= 3tx^2 - 8t^2x +4t^3

b. How should t be chosen so as to maximize V for any given value of x?

c. is there a value of x that maximizes the volume of oil that can be carried?


I've solved parts a and d to this problem and now only b and c remain.

For b, I took the partial derivative of V with respect to x and solved for t, resulting in t=(3/4)x. this is clearly wrong, as the geometry doesn't quite work out. c seems to be dependent on the answer to b. This is in a Microeconomic text book in a section which summarizes the math needed for subsequent chapters. I'm just not sure how to approach this.

Any help would be GREATLY appreciated.
 
jdb said:
A simple way to model the construction of an oil tanker is to start with a large rectangle sheet of steel that is x feet wide and 3x feet long. Now cut a smaller square that is y feet on a side out of each corner of the larger sheet and fold up and weld the sides of the steel sheet to make a traylike structure with no top.

I've shown that the volume can be represented by the equation:

V= 3tx^2 - 8t^2x +4t^3

b. How should t be chosen so as to maximize V for any given value of x?

c. is there a value of x that maximizes the volume of oil that can be carried?
Click to expand...
The squares that are being cut out are \(\displaystyle y\) feet on a side, so stay with that variable instead of t.

The volume is \(\displaystyle lwh = (3x - 2y)(x - 2y)y = 3yx^2 - 8y^2x + 4y^3.\)

The size of the squares to be cut out depend on the size (area) of the rectangular sheet.
Or, the size of the squares are proportionate to the area of the rectangular sheet.
There is a constant ratio of the length of a cut out side of a square to the number of feet
of the width of the rectangular sheet, \(\displaystyle x.\)

For the sake of simplicity, let \(\displaystyle \ \ x = 1\ foot.\)

The equation becomes \(\displaystyle V = 3y - 8y^2 + 4y^3.\)

\(\displaystyle \frac{dV}{dy} = 3 - 16y + 12y^2 = 0\)

The only \(\displaystyle y\)-value that makes sense here is \(\displaystyle \frac{4 - \sqrt{7}}{6} \approx 0.2257\)

Then, to maximize \(\displaystyle V\), \(\displaystyle y \approx 0.2257x.\)

---------------------------------------------------------------------------------------------------------

There is not a value of \(\displaystyle x\) that can maximize the volume of oil, because the sides
of the rectangular sheet can vary. For any given positive \(\displaystyle x\)-value, the maximum
value of the volume depends on \(\displaystyle y\) being equal to about \(\displaystyle 0.2257\) times the number
of feet of \(\displaystyle x.\)
 
I figured it out before I read this, but thanks a ton for the help!!!!!
 
You must log in or register to reply here.
Top