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?
let the original length and width be a, b, and x
abc = 4320
2ab + 2ac + 2bc = 1704 -----> ab + ac + bc = 852
4a + 4b + 4c = 208
a+b+c = 52
new sides: a+1, b+1, c+1
new volume = (a+1)(b+1)(c+1)
= (a+1)(bc + b + c + 1)
= abc + ab + ac + a + bc + b + c + 1
= abc + (ab+ac+bc) + (a+b+c) + 1
= 4320 + 852 + 52 + 1
= .....
sad
urmom
Well, well, well, looks like we have a case of Box and the Magic Inch!
Let's call the length, width, and height of the original box as $l$, $w$, and $h$ respectively.
We know that the volume ($V$) of a rectangular box is given by $V = l \times w \times h$.
And the surface area ($A$) can be calculated using the formula $A = 2lw + 2lh + 2wh$.
Given the volume and surface area of the original box, we have:
$V = 4320$ cubic inches
$A = 1704$ square inches
And since we all know that the sum of the lengths of the edges is $208$ inches, it's time to put on our thinking caps and come up with some equations.
The sum of all the edges of the box is $2(l + w + h)$. So, we have $2(l + w + h) = 208$.
We also know that $A = 2lw + 2lh + 2wh$, so we can rewrite this as $1704 = 2lw + 2lh + 2wh$.
Now, let's flex our math muscles and solve these equations! (Don't worry, they won't bite.)
From the equation $2(l + w + h) = 208$, we can simplify it to $l + w + h = 104$.
Now, let's substitute this value into the equation $1704 = 2lw + 2lh + 2wh$:
$1704 = 2lw + 2lh + 2wh$
$1704 = 2(lw + lh + wh)$
$lw + lh + wh = 852$
Now, let's find the new volume when we increase each side of the box by one inch.
The new volume ($V'$) would be $(l + 1)(w + 1)(h + 1)$.
Expanding this out, we get:
$V' = lwh + lh + wh + lw + l + w + h + 1$
But hang on a second, we already know the values of $l + w + h$ and $lw + lh + wh$ from our earlier calculations!
So, we can substitute these values:
$V' = 4320 + 852 + 1$
$V' = 5173$ cubic inches
Ta-da! The new volume of the box, when each side is increased by one inch, would be $5173$ cubic inches.
Now you've got a box with an upgrade! Enjoy the extra inch!
To solve this problem, we can start by considering the given information.
Let's assume the length, width, and height of the rectangular box are $l$, $w$, and $h$ respectively.
Given that the volume of the box is $4320$ cubic inches, we have the equation:
$$lwh = 4320 \qquad \text{(Equation 1)}$$
It is also given that the surface area of the box is $1704$ square inches. The surface area of a rectangular box can be calculated using the formula:
$$2lw + 2lh + 2wh = 1704 \qquad \text{(Equation 2)}$$
Lastly, we are told that the sum of the lengths of all $12$ edges is $208$ inches. Each edge appears twice when calculating the sum, so we have:
$$2l + 2w + 2h = 208 \qquad \text{(Equation 3)}$$
Now, we need to find the volume of the box when each dimension is increased by one inch, i.e., $l+1$, $w+1$, and $h+1$. Let's represent this new volume as $V_{\text{new}}$.
The volume of the new box can be calculated using the equation:
$$(l+1)(w+1)(h+1) = V_{\text{new}} \qquad \text{(Equation 4)}$$
To find $V_{\text{new}}$, we need to solve the system of equations formed by Equation 1, Equation 2, and Equation 3. Once we find the values of $l$, $w$, and $h$, we can substitute them into Equation 4 to find $V_{\text{new}}$.
Let's proceed to solve the equations to find $l$, $w$, and $h$.