Answer :
Let the width be [tex]$w$[/tex] centimeters. According to the problem, the length is twice the width and the height is [tex]$3$[/tex] centimeters less than the width. Thus, we have:
- Length: [tex]$2w$[/tex]
- Height: [tex]$w - 3$[/tex]
The volume [tex]$V$[/tex] of a rectangular prism is given by
[tex]$$
V = \text{length} \times \text{width} \times \text{height}.
$$[/tex]
We are given that [tex]$V = 120\text{ cm}^3$[/tex], so
[tex]$$
w \cdot (2w) \cdot (w-3) = 120.
$$[/tex]
Simplify the left side:
[tex]$$
2w^2(w-3) = 120.
$$[/tex]
Expanding the expression:
[tex]$$
2w^3 - 6w^2 = 120.
$$[/tex]
Divide the entire equation by [tex]$2$[/tex] to simplify:
[tex]$$
w^3 - 3w^2 = 60.
$$[/tex]
Bring all terms to one side:
[tex]$$
w^3 - 3w^2 - 60 = 0.
$$[/tex]
We now have a cubic equation:
[tex]$$
w^3 - 3w^2 - 60 = 0.
$$[/tex]
This cubic does not factor nicely using simple factoring methods, so we look for a positive real solution. We can estimate the solution by testing values of [tex]$w$[/tex]:
1. For [tex]$w = 5$[/tex]:
[tex]$$
5^3 - 3(5)^2 - 60 = 125 - 75 - 60 = -10.
$$[/tex]
2. For [tex]$w = 6$[/tex]:
[tex]$$
6^3 - 3(6)^2 - 60 = 216 - 108 - 60 = 48.
$$[/tex]
Since the function changes sign between [tex]$w = 5$[/tex] and [tex]$w = 6$[/tex], there is a root in that interval. Testing [tex]$w \approx 5.2$[/tex]:
- Compute [tex]$5.2^3$[/tex]:
[tex]$$
5.2^3 \approx 140.608.
$$[/tex]
- Compute [tex]$3(5.2)^2$[/tex]:
[tex]$$
3(27.04) = 81.12.
$$[/tex]
- Substitute back into the equation:
[tex]$$
140.608 - 81.12 - 60 \approx -0.512.
$$[/tex]
This is very close to [tex]$0$[/tex], indicating that [tex]$w \approx 5.2$[/tex] is a good approximation. Refining further would show that the positive real solution is approximately
[tex]$$
w \approx 5.21 \text{ cm}.
$$[/tex]
Thus, the width of the prism is approximately [tex]$5.21$[/tex] centimeters.
To summarize the dimensions:
- Width: [tex]$w \approx 5.21$[/tex] cm,
- Length: [tex]$2w \approx 10.42$[/tex] cm,
- Height: [tex]$w - 3 \approx 2.21$[/tex] cm.
These dimensions give a volume approximately equal to
[tex]$$
5.21 \times 10.42 \times 2.21 \approx 120 \text{ cm}^3,
$$[/tex]
confirming the solution.
Therefore, the answer for the width in centimeters is approximately
[tex]$$
\boxed{5.21}.
$$[/tex]
- Length: [tex]$2w$[/tex]
- Height: [tex]$w - 3$[/tex]
The volume [tex]$V$[/tex] of a rectangular prism is given by
[tex]$$
V = \text{length} \times \text{width} \times \text{height}.
$$[/tex]
We are given that [tex]$V = 120\text{ cm}^3$[/tex], so
[tex]$$
w \cdot (2w) \cdot (w-3) = 120.
$$[/tex]
Simplify the left side:
[tex]$$
2w^2(w-3) = 120.
$$[/tex]
Expanding the expression:
[tex]$$
2w^3 - 6w^2 = 120.
$$[/tex]
Divide the entire equation by [tex]$2$[/tex] to simplify:
[tex]$$
w^3 - 3w^2 = 60.
$$[/tex]
Bring all terms to one side:
[tex]$$
w^3 - 3w^2 - 60 = 0.
$$[/tex]
We now have a cubic equation:
[tex]$$
w^3 - 3w^2 - 60 = 0.
$$[/tex]
This cubic does not factor nicely using simple factoring methods, so we look for a positive real solution. We can estimate the solution by testing values of [tex]$w$[/tex]:
1. For [tex]$w = 5$[/tex]:
[tex]$$
5^3 - 3(5)^2 - 60 = 125 - 75 - 60 = -10.
$$[/tex]
2. For [tex]$w = 6$[/tex]:
[tex]$$
6^3 - 3(6)^2 - 60 = 216 - 108 - 60 = 48.
$$[/tex]
Since the function changes sign between [tex]$w = 5$[/tex] and [tex]$w = 6$[/tex], there is a root in that interval. Testing [tex]$w \approx 5.2$[/tex]:
- Compute [tex]$5.2^3$[/tex]:
[tex]$$
5.2^3 \approx 140.608.
$$[/tex]
- Compute [tex]$3(5.2)^2$[/tex]:
[tex]$$
3(27.04) = 81.12.
$$[/tex]
- Substitute back into the equation:
[tex]$$
140.608 - 81.12 - 60 \approx -0.512.
$$[/tex]
This is very close to [tex]$0$[/tex], indicating that [tex]$w \approx 5.2$[/tex] is a good approximation. Refining further would show that the positive real solution is approximately
[tex]$$
w \approx 5.21 \text{ cm}.
$$[/tex]
Thus, the width of the prism is approximately [tex]$5.21$[/tex] centimeters.
To summarize the dimensions:
- Width: [tex]$w \approx 5.21$[/tex] cm,
- Length: [tex]$2w \approx 10.42$[/tex] cm,
- Height: [tex]$w - 3 \approx 2.21$[/tex] cm.
These dimensions give a volume approximately equal to
[tex]$$
5.21 \times 10.42 \times 2.21 \approx 120 \text{ cm}^3,
$$[/tex]
confirming the solution.
Therefore, the answer for the width in centimeters is approximately
[tex]$$
\boxed{5.21}.
$$[/tex]
