Maximizing Volume of a Rectangular Box with Given Constraints

[tnutty]

Homework Statement

Johnny needs to make a rectangular box for his physics class project. He has bought P cm of wire and S cm^2 of special paper. He would like to use all the wire (for the 12 edges) and paper (for the 6 sides) to make the box.

What is the largest volume of the box that Johnny can make?
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One could use any method to solve the above problem.
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I just need help getting started.

First for the wire.

I think the perimeter function for the wire is : 4(x+2y) [ check my work please].

We are given S, the area of the paper. So a function for the paper might be, x^2

And we have the volume function for the box : V = xyz

So we have 3 functions :

Perimeter :z = 4(x+2y)
Area : y = x^2
Volume : V = xyz

From there I guess I can substitute. But I know something is wrong.

Any help solving this problem. Hints will be fine. Can I do this with double integrals?

 

[tiny-tim]

Hi tnutty!

Johnny needs to make a rectangular box for his physics class project. He has bought P cm of wire and S cm^2 of special paper. He would like to use all the wire (for the 12 edges) and paper (for the 6 sides) to make the box.

What is the largest volume of the box that Johnny can make?
…
Perimeter :z = 4(x+2y)
Area : y = x^2
Volume : V = xyz

Yes, the volume is right, but the perimeter and area aren't.

Try again … what are the perimeter and area of a reactangular box with sides x y and z ?

 

[tnutty]

Perimeter : 2(2x + 2y) + 2(2z) = 4x + 4y + 4z
Area of a rectangle box = 2(xy) + 2(xz) + 2(yz) = 2xy + 2xz + 2yz
V = xyz

What do you think?

Also how do you think we could incorporate the variable P and S ?

 

[tiny-tim]

Hi tnutty!

(just got up :zzz: …)

Perimeter : 2(2x + 2y) + 2(2z) = 4x + 4y + 4z
Area of a rectangle box = 2(xy) + 2(xz) + 2(yz) = 2xy + 2xz + 2yz
V = xyz

That's correct (though a little long-winded).

Hint: what is Area times z ? ​

 

[tnutty]

Area times z is also volume no? And what is meant by long-winded ?

So
V = xyz
V = A * z
A = 2x(y+z) + 2yz
P = 4(x+y+z)

?

 

[tiny-tim]

Area times z is also volume no? And what is meant by long-winded ?

So
V = xyz
V = A * z
A = 2x(y+z) + 2yz
P = 4(x+y+z)

?

No, V is not Az

A = 2(xy + xz + yz), so what is Az in terms of P and V ?

(and "long-winded" was referring to the intermediate stage with all those brackets)

 

[tnutty]

No, V is not Az

A = 2(xy + xz + yz), so what is Az in terms of P and V ?

(and "long-winded" was referring to the intermediate stage with all those brackets)

I thought Az was the volume? Since x*y = Area, then V = Az. ?

I am not sure about the second question.

A = 2(xy+xz+yz ) then

A*z = 2z(xy+xz+yz) ?

 

[tiny-tim]

I thought Az was the volume? Since x*y = Area, then V = Az. ?

You're using "area" to mean two different things.

xy is the area of one face of the box.

There are three different sizes of face, and A is the sum of the areas of all the faces.

I am not sure about the second question.

A = 2(xy+xz+yz ) then

A*z = 2z(xy+xz+yz) ?

Yes … and what is that in terms of P and V ?

 

[tnutty]

Yes … and what is that in terms of P and V ?

A*z = 2z(xy+xz+yz)

V = xyz
x = V/yz


P/4 = x+y+z
P/4 - x - y = z


A*z = 2z(xy+xz+yz)

A( P/4 - x - y ) = 2(P/4 - x -y) ( V/z + V/y + yz )

is that right?

You're using "area" to mean two different things.

xy is the area of one face of the box.

There are three different sizes of face, and A is the sum of the areas of all the faces.

For a rectangular box, we could compute the area of one size and multiply by its depth
to get the volume no? I am confused why it isn't. Can you take a look at this picture, http://img64.imageshack.us/img64/415/31659034.png" .

The integration of an area over a region is the object's volume right? So
the way I thought about it is that we compute the area of a strip of the box( the red strip in the picture), This strip is dz. And since the area does not change over dz, we can pull the area out and the integral reduces to Area * Z . Thus volume is Area * Z, or Area*Depth?