SOLUTION: Find the dimensions of the box described. The length is 4 inches more than the width. The width is 3 inches more than the height. The volume is 480 cubic inches.

Question 1127661: Find the dimensions of the box described.
The length is 4 inches more than the width. The width is 3 inches more than the height. The volume is 480 cubic inches.

Found 2 solutions by FrankM, greenestamps:
Answer by FrankM(1040) (Show Source): You can put this solution on YOUR website!
H=X W=X+3 L=X+7
LWH = X^3+10X^2+21X=480 so

X^3+10X^2+21X-480=0

Using a bit of guess and check, we easily find X=5

(Note, X has to be positive, and at 1, we get 32-480, very negative. Jumping to 3, we have 27+90+63-480= -300, so we jump to 5 and see 125+250+105-480=0)

5 * 8 * 12 =480
Answer by greenestamps(13200) (Show Source): You can put this solution on YOUR website!

The solution by tutor FrankM is probably right in saying that the solution is found more easily with guess-and check than with formal algebra.

But I would do the guess-and-check work differently than what he shows.

Without expanding the expression, the equation for the volume of the box is

x(x+3)(x+7)=480

Since the product is 480, one of the factors must be 5, or perhaps a multiple of 5.

So try x = 5, and it works.

So the dimensions are 5, 8, and 12.
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