Find the most economical dimensions of a closed rectangular box [..]

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SUMMARY

The discussion focuses on determining the most economical dimensions of a closed rectangular box with a volume of 8 cubic units, considering varying material costs for different sides. The cost structure includes 5 per square unit for the top and bottom, 2 for the front and back, and 5 for the other two sides. The solution involves using the Lagrange Multiplier method to derive equations for the dimensions, specifically focusing on the relationships between length (x), width (y), and height (z) through cost equations and volume constraints.

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s3a
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"Find the most economical dimensions of a closed rectangular box [. . .]"

Homework Statement


Find the most economical dimensions of a closed rectangular box of volume 8 cubic units if the cost of the material per square unit for (i) the top and bottom is 5, (ii) the front and back is 2 and (iii) the other two sides is 5.

(a) What is the length of the vertical edge?
(b) What is the length of the horizontal front and back edge?
(c) What is the length of the horizontal side edge?


Homework Equations


Lagrange Multiplier Method or equivalent.


The Attempt at a Solution


The Lagrange Multiplier method is not required for this problem so if there is an easier way please notify me of it as well as why it's easier. My work is attached but I'm stuck at the step of figuring out the variables/dimensions.

Any help would be GREATLY appreciated!
Thanks in advance!
 

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It would make more sense if you would specify what x, y, and z represent at the start.
I assume you mean that x and y are "length" and "width" and z is "height".

In any case, yes, the volume satifies xyz= 8 and the "cost" is 10xy+ 4xz+ 10yz.

You can write each of your erquations as
10y+ 4z= -\lambda yz
10x+ 10z= -\lambda xz and
4x+ 10z= -\lambda xy

Since the value of \lambda is not important to the solution, I like to eliminate it first by dividing one equation by another:
If you divide the first equation by the second you get
\frac{5y+ 2z}{5x+ 5z}= \frac{y}{x}
and if you divide the first equation by the third you get
\frac{5x+ 2z}{2x+ 5z}= \frac{z}{x}
Those, together with xyz= 8 give you three equations to solve for x, y, and z.
 


s3a said:

Homework Statement


Find the most economical dimensions of a closed rectangular box of volume 8 cubic units if the cost of the material per square unit for (i) the top and bottom is 5, (ii) the front and back is 2 and (iii) the other two sides is 5.

(a) What is the length of the vertical edge?
(b) What is the length of the horizontal front and back edge?
(c) What is the length of the horizontal side edge?


Homework Equations


Lagrange Multiplier Method or equivalent.


The Attempt at a Solution


The Lagrange Multiplier method is not required for this problem so if there is an easier way please notify me of it as well as why it's easier. My work is attached but I'm stuck at the step of figuring out the variables/dimensions.

Any help would be GREATLY appreciated!
Thanks in advance!

Your attachment is so messy that it is unreadable; please be more careful when posting here.

RGV
 

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