SOLUTION: SUPPOSE YOU ARE BUILDING A STORAGE BOX OF VOLUME 4368 in3. The length of the box will be 24 in. The height of the box will be 1 in more than its width, Find the height and the wid

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Question 196384This question is from textbook FCAT DAILY SKILLS AND PRACTICE WORKBOOK
: SUPPOSE YOU ARE BUILDING A STORAGE BOX OF VOLUME 4368 in3. The length of the box will be 24 in. The height of the box will be 1 in more than its width, Find the height and the width.
I know v=lwh
v=4368in3
l=24
w=x
h=x+1

(24)(x)(x + 1) = 4368in3
(24 + x)(x + 1) = 4368in3
This question is from textbook FCAT DAILY SKILLS AND PRACTICE WORKBOOK

Found 2 solutions by scott8148, Edwin McCravy:
Answer by scott8148(6628)   (Show Source): You can put this solution on YOUR website!
(24)(x)(x+1)=4368in3

dividing by 24 ___ x^2+x = 182

subtracting 182 ___ x^2 + x - 182 = 0

factoring ___ (x+14)(x-13) = 0

width = 13

height = 14

Answer by Edwin McCravy(20056)   (Show Source): You can put this solution on YOUR website!
Edwin's answer:
SUPPOSE YOU ARE BUILDING A STORAGE BOX OF VOLUME 4368 in3. The length of the box will be 24 in. The height of the box will be 1 in more than its width, Find the height and the width.
I know v=lwh
v=4368in3
l=24
w=x
h=x+1
(24)(x)(x + 1) = 4368in3
(24 + x)(x + 1) = 4368in3
The inches cubed is confusing me because I know I have to factor and substitute 0 to solve for x but I am lost on this one.

Your first equation is correct but the second one is wrong.
It should be 24x not (24 + x), you're multiplying 24 by x. You're
not adding them.

Maybe this will help with your "inch cubed" confusion:

Below is one INCH, its LENGTH is 1 inch raised to the
1 power. You can think of it as 1 INCH1.



Below is one SQUARE INCH, its AREA is found by 
multiplying 1 inch length by 1 inch width, which is 1 
INCH SQUARED, or 1 INCH2, or 1 SQUARE INCH.



Below is one CUBIC INCH, its VOLUME is found by 
multiplying 1 inch length by 1 inch width by 1 inch 
height, which is 1 INCH CUBED, or 1 INCH3.



What you really have in your first equation is:

(24 in)(x in)(x + 1 in) = 4368 in3

See the three " in " factors on the left? A unit of length,
such as an inch, is just like a number. It can be raised to powers 
the same as numbers can.

When you raise an inch to the first power, you have a measurement of
LENGTH.

When you raise an inch to the second power, you have a measurement of
AREA.

When you raise an inch to the third power, you have a measurement of
VOLUME.

You can start by putting all the inches in your equation on both
sides like this:

(24 in)(x in)(x + 1 in) = 4368in 3

Then the inches on the left side are multiplied three times, so
we can write the left side as 

(24)(x)(x + 1)in3 = 4368in3

Now you can divide both sides by in3, and you
just have the equation:

(24)(x)(x + 1) = 4368    with no inches in it at all.

Lets write (24)(x) as 24x, OK?

24x(x + 1) = 4368

We could go ahead and distribute, but what we should do 
to make things easier instead is to divide both sides by 24,
since 4368 can be divided by 24 to give 182, so we have:

x(x + 1) = 182

Now you distribute:

x2 + x = 182

Get 0 on the right by subtracting 182
from both sides:

x2 + x - 182 = 0

To factor that we have to think of a pair
of inegers that have product 182 and difference 1.
So they will have to be very near the square root
of 182.  The square root of 182 is 13.49073756,
so the two closest integers to that are 13 and 14.
And they do differ by 1 and have product 182.
So we factor it this way:

(x - 13)(x + 14) = 0

Set each factor = 0

x - 13 = 0
     x = 13

x + 14 = 0
     x = -14

We discard the negative answer.  So the width is x inches
or 13 inches, and the height is x + 1 or 13 + 1 or 14 inches.

Edwin

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