How Do You Find the Dimensions of a Prism with a Volume of 42 cm³?

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To find the dimensions of a rectangular prism with a volume of 42 cm³, the height is defined as x-1, width as x-2, and length as x+3. The equation derived from the volume is 42 = (x-1)(x-2)(x+3), which simplifies to a cubic equation. The solution suggests that x equals 4, leading to dimensions of 3 cm, 2 cm, and 7 cm for the prism. Graphical methods or trial and error can be effective for solving such cubic equations, focusing on positive, real solutions. Understanding the conditions for valid dimensions is crucial, as negative or zero lengths are not meaningful.
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Homework Statement


Heres the question I'm tackling:
<b> the height of a prism is x-1, width x-2, length x+3 and the volume is 42cm cubed. Find the dimensions</b>


Homework Equations


Quadratic formula, common factor, family of functions



The Attempt at a Solution


I know that 7, 2 and 3 are multiples of 42 but i do not know how to show my work... I have this so far...
42=(x-1)(x-2)(x+3)
42=(x-1)(x^2-x-6)
42=(x^3-x^2-6x-x^2+x+6)
42=(x^3-2x^2-5x+6)

and I do not know if I am doing things right...
 
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Is that a rectangular, triangular, or what kind of prism..
 
Rectangular prism... sorry for the misleading question
 
42=(x-1)(x-2)(x+3)
42=(x-1)(x^2-x-6)

! That should be +x. Makes rest of your calculations wrong.

I got it to be
(x^3)-7x-36=0
or
(x^3)-7x+6=42

And that's all I got so far..
---
PS, 42 cubed? That adds another dyanmic..
Maybe (x^3)-7x+6=(42^3)
?
 
HEY, using simple random "pluggin in random numbers", I think the "x" is 4..
Therefore, the sides would be
(4-1), (4-2), (4+3)
or
3, 2, 7

Now, someone else explain an algebraic way of doing this:D
 
Ya well I am not really sure how to do all this... but that's a good theory.. It is a rectangular prism and 42 is 42cm^2. Although, I'm not even sure if I'm doing all this right
 
Yo ill tinks That works! THANKS
 
Although through the pluggin theory i need to know how to use the right methods to get to 4
 
  • #10
Say you graphed it and that was the only solution
 
  • #11
4 is the correct answer I just do not know how to get to it... My teacher vaguely told me anything about questions like this.
 
  • #12
Graph..
y=(x^3)-7x-36

See where the 0 is..at 4.. What does that mean logically I don't know.:D
 
  • #13
To solve the equation in this case, you could use the graphical method; i.e. find the roots of the equation x^3-7x-36=0. However, since this is a real problem, looking for a real, positive length, there are certain conditions that the solution must satisfy. Firstly, we are not interested in complex solutions. We note from above that x cannot be negative, or zero, (since then at least one of the sides has negative length, which has no meaning), and x cannot be 1 or 2 (since in each case this gives us a side with zero length)

So, simply start plugging in values starting with 3; you won't have to put many values into obtain the result!
 
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  • #14
I don't have a graphing calculator...
 
  • #16
@/@ said:
Some search

you can use the ruffini rule and the rational root theorem

http://en.wikipedia.org/wiki/Ruffini's_rule

look the section
Polynomial root-finding

IMHO, it is not necessary to learn these methods of finding roots to a cubic equation for these simple cases. At least one of the roots of problems of this level will be a small number, so "guessing" a solution is enough. Of course, if one wanted to find all the roots, and has found one a, say, using the "trial and error" method, then the equation can be reduced to a quadratic by dividing (x-a) into it. The remaining polynomial can be solved using the usual quadratic formula.
 
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