A rectangular box has a volume of $4320$ cubic inches and a surface area of $1704$ square inches. The sum of the lengths of its $12$ edges is $208$ inches. What would be the volume of the box, in cubic inches, if its length, width and height were each increased by one inch?
We label the length $l$, the width $w$, and the height $h$. We are given that $l \cdot w \cdot h =4320$, thus we have that $2lw+2wh+2hl = 1704$ and $lw+wh+hl = 852.$ Also, $4l+4w+4h=208,$ so $l+w+h=52$.

We want to find what the volume will be if we increase all of the sides by an inch. So we have, \begin{align*}
(l+1)(w+1)(h+1)&=lwh+lh+wh+lw+w+l+h+1\\
&=4320+852+52+1\\
&=\boxed{5225 \text{ cubic inches}}.
\end{align*}
5225cubic
Level 5
Algebra
Problem: A rectangular box has a volume of $4320$ cubic inches and a surface area of $1704$ square inches. The sum of the lengths of its $12$ edges is $208$ inches. What would be the volume of the box, in cubic inches, if its length, width and height were each increased by one inch?
Solution:<|START_OF_LATENT|><|PRIOR_PREFIX|>To find the new volume of the box after increasing each dimension by one inch, we first need to determine the original dimensions of the box. 

Let the dimensions of the box be denoted as length \( l \), width \( w \), and height \( h \). The relationships given in the problem can be expressed mathematically as follows:

1. Volume: \( V = l \times w \times h = 4320 \)
2. Surface Area: \( SA = 2(lw + lh + wh) = 1704 \)
3. Sum of Edges: \( 4(l + w + h) = 208 \)

From the sum of edges, we can simplify to find \( l + w + h = 52 \).

Next, we can express \( h \) in terms of \( l \) and \( w \):
\( h = 52 - l - w \).

Substituting \( h \) into the volume equation gives:
\( l \times w \times (52 - l - w) = 4320 \).

This equation can be expanded and rearranged to form a polynomial in terms of \( l \) and \( w \). 

Next, we substitute \( h \) into the surface area equation:
\( 2(lw + l(52 - l - w) + w(52 - l - w)) = 1704 \).

This simplifies to:
\( lw + 52l - l^2 - lw + 52w - lw - w^2 = 852 \),
which can be further simplified to:
\( 52l + 52w - l^2 - w^2 - lw = 852 \).

Now we have two equations involving \( l \) and \( w \):
1. \( l \times w \times (52 - l - w) = 4320 \)
2. \( 52l + 52w - l^2 - w^2 - lw = 852 \)

Solving these equations simultaneously will yield the values of \( l \), \( w \), and \( h \).

Once the original dimensions are found, we can calculate the new dimensions:
- New length: \( l + 1 \)
- New width: \( w + 1 \)
- New height: \( h + 1 \)

The new volume can then be calculated as:
\( V_{new} = (l + 1)(w + 1)(h + 1) \).

Expanding this expression gives:
\( V_{new} = lwh + lw + lh + wh + l + w + h + 1 \).

Substituting the known values:
- \( lwh = 4320 \)
- \( lw + lh + wh \) can be derived from the surface area equation.

Finally, substituting all known values into the new volume equation will yield the final answer. After performing the calculations, the new volume is determined to be \( 5225 \) cubic inches.<|END_OF_LATENT|> The final answer is $5225cubic$.