Maximum volume word problem

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warwick

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A rectangular piece of cardboard measuring 20 inches by 32 inches is to be made into a box with an open top by cutting equal size squares from each corner and folding up the sides. Let x represent the length of a side of each such square. What is the maximum volume of this box?

I know you can find the solution through Calculus by calculating the first and second derivates, but how can you solve this using only Pre-Calculus?

The volume equation I found was:

4x^3 - 104x^2 +640x.

I can't believe I'm having a mental bloc on this. Thanks for any help.
 
warwick said:
The volume equation I found was 4x^3 - 104x^2 +640x. I can't believe I'm having a mental bloc on this.
Click to expand...
I agree with your equation but, off the top of my head, I can't see how you're supposed to do this without calculus, either. I mean, if this were a second-degree equation, sure: you'd find the vertex of the parabola, instead of finding the zero of the first derivative. But a cubic...?

Here's hoping somebody more clever comes along with something useful....

Eliz.
 
stapel said:
warwick said:
The volume equation I found was 4x^3 - 104x^2 +640x. I can't believe I'm having a mental bloc on this.
Click to expand...
I agree with your equation but, off the top of my head, I can't see how you're supposed to do this without calculus, either. I mean, if this were a second-degree equation, sure: you'd find the vertex of the parabola, instead of finding the zero of the first derivative. But a cubic...?

Here's hoping somebody more clever comes along with something useful....

Eliz.
Click to expand...

I found several discrepancies with the Sample Exam and the Solution Sheet, as well as incorrect solutions, and emailed the professor; he did forward it to everyone. I did find a few more, too, after I emailed him.
 
The only thing I could find online was an article from "Ask Dr. Math" which confirmed my suspicions: If you can't use calculus, then you have to work from the graph.

If you've got a graphing calculator, great: plug in your "volume" function and ask the calculator to show you the "max" on the graph. Otherwise, graph by hand.

Hint: The x-value that looks like it gives you the maximum volume is the maximizing value. The x-value and thus the volume happen, in this particular case, to be nice whole numbers.

Eliz.
 
Hello, warwick!

A rectangular piece of cardboard measuring 20 inches by 32 inches is to be made into a box
with an open top by cutting equal size squares from each corner and folding up the sides.
Let x represent the length of a side of each such square.
What is the maximum volume of this box?

I know you can find the solution through Calculus by calculating the first and second derivates,
but how can you solve this using only Pre-Calculus? \(\displaystyle \;\;\) . . . I'm not sure

The volume equation I found was: \(\displaystyle \,V\:=\:4x^3\,-\,104x^2\,+\,640x\;\) . . . correct!
Click to expand...
I suppose we could try graphing the cubic and estimate the maximum . . .

For x-intercepts, set the function equal to 0 and solve:
\(\displaystyle \;\;4x(x^2\,-\,26x\,+\,160)\:=\:0\;\;\Rightarrow\;\;4x(x\,-\,10)(x\,-\,16)\:=\:0\)

The x-intercepts are: 0, 10, 16 . . . the graph looks like this:
Code: 
        |   ***
        | *     *
        |*       *           *
        |
    ----*---------*---------*--
        |        10          16
       *|          *       *
        |           *     *
        |             ***
From the physical constraints of the problem, we know that: \(\displaystyle \,0\,<\,x\,<\,10\)
So we are concerned with only the first "hump' of the graph.

And we can try different values of x to find that maximum point.

As Eliz. pointed out, it comes out nicely this time: \(\displaystyle \,x\,=\,4\)
 
soroban said:
Hello, warwick!

A rectangular piece of cardboard measuring 20 inches by 32 inches is to be made into a box
with an open top by cutting equal size squares from each corner and folding up the sides.
Let x represent the length of a side of each such square.
What is the maximum volume of this box?

I know you can find the solution through Calculus by calculating the first and second derivates,
but how can you solve this using only Pre-Calculus? \(\displaystyle \;\;\) . . . I'm not sure

The volume equation I found was: \(\displaystyle \,V\:=\:4x^3\,-\,104x^2\,+\,640x\;\) . . . correct!
Click to expand...
I suppose we could try graphing the cubic and estimate the maximum . . .

For x-intercepts, set the function equal to 0 and solve:
\(\displaystyle \;\;4x(x^2\,-\,26x\,+\,160)\:=\:0\;\;\Rightarrow\;\;4x(x\,-\,10)(x\,-\,16)\:=\:0\)

The x-intercepts are: 0, 10, 16 . . . the graph looks like this:
Code: 
        |   ***
        | *     *
        |*       *           *
        |
    ----*---------*---------*--
        |        10          16
       *|          *       *
        |           *     *
        |             ***
From the physical constraints of the problem, we know that: \(\displaystyle \,0\,<\,x\,<\,10\)
So we are concerned with only the first "hump' of the graph.

And we can try different values of x to find that maximum point.

As Eliz. pointed out, it comes out nicely this time: \(\displaystyle \,x\,=\,4\)
Click to expand...

Thanks. The answer on the solution sheet is 4.15, but I used my TI-89 Titanium and the maximum it gave me was 4, as well. How did the professor get 4.15 then?
 
warwick said:
How did the professor get 4.15 then?
Click to expand...
I have no clue how he got that. It's clearly a wrong answer.

Eliz.
 
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