PROBLEM SOLVING INVOLVING POLYNOMIAL EQUATION

fayleeee

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I need your help please : A shipping container in the shape of a rectangular solid must have a volume of 84 cubic meters. The client tells the manufacturer that, because of the contents, the length of the container must be one meter longer than the width, and the height must be one meter greater than the twice the width. What should the dimensions of the container be?
 
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I need your help please : A shipping container in the shape of a rectangular solid must have a volume of 84 cubic meters. The client tells the manufacturer that, because of the contents, the length of the container must be one meter longer than the width, and the height must be one meter greater than the twice the width. What should the dimensions of the container be?

You have three unknowns that you need to calculate. Name those first. Let:

the length of the container = L

the width of the container = W

the height of the container = H

You know the volume (V) \(\displaystyle \ \ \to \ \ \)V = L * W * H = 84

What are conditions are given? continue....
 
In general, the way to solve a system of equations (assuming that is possible) is through substitution. The fact that you will end up with a cubic equation does not alter that general rule.
 
hi! i somehow have the same problem that my teacher gave me earlier. i was wondering, did i do it right?

V= 84
L=1+X
W=X
H= 1+2X
( X=?)

V=LWH
84= X(1+X)(1+2X)
84=(X+X^2)(1+2X)
84=X-X+2X^3
84/2= 2X^3/2
42=X^3
√42=X
X= 3.476--------> X≈3

and then i substituted the value of x to the L,W and H resulting to 4,3 and 7 respectively.
 
Last edited:
V=LWH
84= X(1+X)(1+2X)
84=(X+X^2)(1+2X)
84=X-X+2X^3 WHERE DID THE MINUS SIGN COME FROM?
84/2= 2X^3/2
42=X^3
√42=X DID YOU TAKE THE CUBE ROOT OR THE SQUARE ROOT?

I think you meant [MATH]\sqrt[3]{42}[/MATH]X= 3.476--------> X≈3

and then i substituted the value of x to the L,W and H resulting to 4,3 and 7 respectively.
Hit the click to expand button for for a better view of my comments on your work.

You did the substitutions correctly. Good for you, but then messed up the algebra.

By any chance are you studying the rational root theorem?
 
Hit the click to expand button for for a better view of my comments on your work.

You did the substitutions correctly. Good for you, but then messed up the algebra.

By any chance are you studying the rational root theorem?
hi, on this part "X-X+2X^3" i got the negative sign when i was applying the foil method. and yes i meant cube root there sorry i cant find the symbol of it hehe im still quite confused on how to use this site
 
in the right hand side of 84=(X+X^2)(1+2X) all the terms are positive!
So how did you get a -x in 84=X-X+2X^3????

If x=3.476 why would you choose to use 3 for x???? It is a amazing that 4*3*7 happens to be 84. That should have shown you that you had the wrong answer. Each of the numbers you used, 4,3 and 7 are UNDERESTIMATES so how can 4*3*7 = 84???!!! 4*3*7 should be under 84!!
 
in the right hand side of 84=(X+X^2)(1+2X) all the terms are positive!
So how did you get a -x in 84=X-X+2X^3????

If x=3.476 why would you choose to use 3 for x???? It is a amazing that 4*3*7 happens to be 84. That should have shown you that you had the wrong answer. Each of the numbers you used, 4,3 and 7 are UNDERESTIMATES so how can 4*3*7 = 84???!!! 4*3*7 should be under 84!!
AAAAAAAAA I JUST REALIZED THAT AH THIS IS SO EMBARASSING. I WILL CHANGE MY ANSWER, THANK YOU SO MUCH
 
AAAAAAAAA I JUST REALIZED THAT AH THIS IS SO EMBARASSING. I WILL CHANGE MY ANSWER, THANK YOU SO MUCH
If you want to avoid errors, do it like this

[MATH]x(1 + x)(1 + 2x) = 84 \implies\\ (x + x^2)(1 + 2x) = 84 \implies\\ x(1 + 2x) + x^2(1 + 2x) = 84 \implies \\ x + 2x^2 + x^2 + 2x^3 = 84 \implies \\ 2x^3 + 3x^2 + x - 84 = 0.[/MATH]
Now you have a cubic equation in standard form. There are at least four ways to solve a cubic. The question is which do you know? That is why I asked about the rational root method.

As for the cube root symbol, do not worry to0 much about that. But if you know about fractional exponents, you have a solution.

By the way, your solution was correct BY ACCIDENT. Now let's get it the right way.

Oh, and thank you for showing your work.
 
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