Solving Problems with Advanced Functions

[maeveoneill]

Question: The height. length, and width of a small box are consecutive integers with the height being the smallest of hte three dimensions. if the length and width are increased by 1cm each and the height is doubled. then the volume is increased by 120cm³. Find the dimension of the original box.

Answer: 3cm, 4cm,5cm

The questions is from my grade 12 advanced functions and introductory calculus course. The answer was taken from the back of the book. However, I do not know how to get it. Any steps would be greatly appreciated and as quickly as possible please.

I have so far come up with the equations v=h³ +3h² +2h (for the small box) and v=2h³ +10h² +12h -120 (for the large box) Where do i go from there?

Thanks!!

 

[stapel]

How did you get your second equation?

Please reply with your reasoning. Thank you.

Eliz.

P.S. Please post any further algebra questions to one of the "algebra" categories. Thank you.

 

[TchrWill]

The initial box is x by (x + 1) by (x + 2)
The second box is 2x by (x + 2) by (x + 3).

Then, 2x(x + 2)(x + 3) - x(x + 1)(x + 2) = 120.

Expand, simlify and solve for x.

 

[soroban]

Hello, maeveoneill!

I made a modification on your set-up . . .

The height. length, and width of a small box are consecutive integers
with the height being the smallest of hte three dimensions.
If the length and width are increased by 1cm each and the height is doubled,
then the volume is increased by 120cm³.
Find the dimension of the original box.
[Answer: 3cm, 4cm, 5cm]

I have so far come up with: V₁ = h³ +3h² +2h (for the small box)

and V₂ = 2h³ +10h² +12h (for the large box). . . . . good!
.

We are told that V₂ is 120 more than V₁.

We have: .2h³ + 10h² + 12h .= .h³ + 3h² + 2h + 120

. . . . . . h³ + 7h² + 10h - 120 .= .0

. . . . . .(h - 3)(h² + 10h + 40) .= .0

Hence, h = 3.

Therefore, the dimensions are 3 x 4 x 5 cm.

. . . [Edit: Too fast for me, TchrWill!]