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?
To solve this problem, we can start by setting up equations using the given information.
Let's assume that the length, width, and height of the box are $l$, $w$, and $h$ respectively.
From the given information, we have the following equations:
1) The volume of the box is $4320$ cubic inches:
\[l \times w \times h = 4320\]
2) The surface area of the box is $1704$ square inches:
\[2lw + 2lh + 2wh = 1704\]
3) The sum of the lengths of the $12$ edges is $208$ inches:
\[4(l + w + h) = 208\]
Now, let's consider increasing the length, width, and height of the box by one inch. The new dimensions would be $(l + 1)$, $(w + 1)$, and $(h + 1)$ respectively.
The volume of the new box can be found using the equation:
\[(l + 1) \times (w + 1) \times (h + 1)\]
Substituting the value of $l \times w \times h$ from equation (1) into the above equation, we get:
\[(l + 1) \times (w + 1) \times (h + 1) = 4320 + (l + 1) \times (w + 1) \times (h + 1) - l \times w \times h\]
Expanding the right side of the equation, we have:
\[(l + 1) \times (w + 1) \times (h + 1) = 4320 + (lw + l + w + 1) \times (h + 1) - l \times w \times h\]
The terms $lw + l + w + 1 - l \times w \times h$ can be simplified using the given equations (1) and (2).
From equation (1), we have $lw \times h = 4320$, so $lw = \frac{4320}{h}$.
From equation (2), we have $2lw + 2lh + 2wh = 1704$. Substituting $lw = \frac{4320}{h}$, we get:
\[2 \times \frac{4320}{h} + 2lh + 2wh = 1704\]
Simplifying the equation, we have:
\[\frac{8640}{h} + lh + wh = 852\]
Multiplying both sides of the equation by $h$ to eliminate the denominator, we have:
\[8640 + l \times h^2 + w \times h^2 = 852 \times h\]
Rearranging terms, we get:
\[l \times h^2 + w \times h^2 - 852 \times h + 8640 = 0\]
We can now substitute $lh = \frac{4320}{w}$ into this equation:
\[\frac{4320}{w} \times h + w \times h^2 - 852 \times h + 8640 = 0\]
Rearranging terms again, we have:
\[\frac{4320}{w} \times h + (w \times h^2 - 852 \times h + 8640) = 0\]
Simplifying this equation gives us:
\[4320h + w(w \times h - 852) = 0\]
Now, let's go back to the equation for the volume of the new box:
\[(l + 1) \times (w + 1) \times (h + 1) = 4320 + (lw + l + w) \times (h + 1) - l \times w \times h\]
Substituting the values obtained from the above equation, we have:
\[(l + 1) \times (w + 1) \times (h + 1) = 4320 - (4320h + w(w \times h - 852))\]
Expanding this equation gives us:
\[(l + 1) \times (w + 1) \times (h + 1) = 4320 - 4320h - w(w \times h - 852)\]
Substituting the value of $4320h + w(w \times h - 852)$ obtained from the previous equation, we get:
\[(l + 1) \times (w + 1) \times (h + 1) = 4320 - 4320h - \left(\frac{4320h + w(w \times h - 852)}{h}\right)\]
Simplifying this equation gives us:
\[(l + 1) \times (w + 1) \times (h + 1) = 4320 - 4320h - \left(\frac{4320h + w^2h - 852w}{h}\right)\]
Further simplification gives us:
\[(l + 1) \times (w + 1) \times (h + 1) = 4320 - 4320h - \left(\frac{4320h + w^2h - 852w}{h}\right)\]
Now, we can simplify the terms by distributing the factors:
\[(lwh + lw + lh + wh + l + w + h + 1) = 4320 - 4320h - \left(\frac{4320h + w^2h - 852w}{h}\right)\]
From equation (3), we know that $4(l + w + h) = 208$, so $l + w + h = \frac{208}{4}$, which simplifies to $l + w + h = 52$.
Substituting this value into the equation, we have:
\[(lwh + lw + lh + wh + 52 + 1) = 4320 - 4320h - \left(\frac{4320h + w^2h - 852w}{h}\right)\]
Simplifying this equation gives us:
\[(lwh + lw + lh + wh + 53) = 4320 - 4320h - \left(\frac{4320h + w^2h - 852w}{h}\right)\]
We have now obtained an equation for the volume of the new box. It can be simplified further, but it is quite complex and doesn't seem to provide a direct solution.
Therefore, it is not possible to determine the exact volume of the box when its length, width, and height are each increased by one inch using the given information.
However, we can make an estimation by approximating the volume using the initial values. Since we know the volume of the original box is $4320$ cubic inches, we can use this as an estimate for the new volume when the dimensions are increased by one inch.
So, the estimated volume of the box would be $4320$ cubic inches if its length, width, and height were each increased by one inch.
Let's denote the length, width, and height of the rectangular box as $l$, $w$, and $h$, respectively.
We know that the volume of the box is given by $V = lwh$, and the surface area is given by $A = 2lw + 2lh + 2wh$. We are also given that the sum of the lengths of its edges is $208$ inches: $l + w + h + l + w + h + l + w + h + l + w + h = 208$.
From this, we can solve for the length, width, and height of the box:
$4(l + w + h) = 208$
$l + w + h = 52$
Now, let's solve for the length, width, and height.
We have the equation $4320 = lwh$ and $1704 = 2lw + 2lh + 2wh$.
We can solve these equations simultaneously to find the values of $l$, $w$, and $h$.
From the equation $l + w + h = 52$, we can rearrange it as $l + w = 52 - h$.
Substituting this into the equation $1704 = 2lw + 2lh + 2wh$, we get:
$1704 = 2l(52 - h) + 2lh + 2wh$
$1704 = 104l - 2lh + 2lh + 2wh$
$1704 = 104l + 2wh$
Now, we can substitute the value of $l$ in terms of $w$ and $h$ into the above equation:
$1704 = 104(52 - h) + 2wh$
$1704 = 5408 - 104h + 2wh$
$1704 - 5408 = -104h + 2wh$
$-3704 = -104h + 2wh$
Dividing by 2, we get:
$-1852 = -52h + wh$
Rearranging this equation:
$wh - 52h -1852 = 0$
Now, let's solve this quadratic equation for $w$:
$w = \frac{52 \pm \sqrt{(-52)^2 - 4(1)(-1852)}}{2(1)}$
$w = \frac{52 \pm \sqrt{2704 + 7408}}{2}$
$w = \frac{52 \pm \sqrt{10112}}{2}$
$w = \frac{52 \pm 100.56}{2}$
Since we know that the width of the box cannot be negative, we take the positive value:
$w = \frac{52 + 100.56}{2}$
$w = \frac{152.56}{2}$
$w = 76.28$
Now, substituting the value of $w$ back into the equation $l + w + h = 52$:
$l + 76.28 + h = 52$
$l + h = -24.28$
$l = -24.28 - h$
Since the length of the box cannot be negative, we need to discard this negative solution. Therefore, we can ignore this case.
So, the width of the box is approximately $76.28$ inches.
Now, let's find the value of $l$ using the given equation $l + w + h = 52$.
$l + 76.28 + h = 52$
$l + h = -24.28$
$l = -24.28 - h$
Substituting $w = 76.28$ into the equation $1704 = 2lw + 2lh + 2wh$, we can solve for $h$:
$1704 = 2l(76.28) + 2l(-24.28) + 2(76.28)h$
$1704 = 152.56l - 48.56l + 152.56h$
$1704 + 45.56l = 152.56h$
Simplifying further:
$1704 + 45.56(-24.28-h) = 152.56h$
$1704 - 1110.97 - 45.56h = 152.56h$
$593.03 = 198.12h$
Dividing both sides by 198.12:
$h = \frac{593.03}{198.12}$
$h \approx 2.995$
Since the height cannot be a fraction, we need to round it to the nearest whole number. Therefore, $h = 3$.
Now that we have the values for $l$, $w$, and $h$, we can calculate the original volume of the box:
$V = lwh = (76.28)(-24.28)(3)$
$V \approx -5589.13$
The original volume of the box is approximately $-5589.13$ cubic inches.
looks like we have 3 unknowns :
l, w, and h for length , width and height
given: lwh = 4320 **
2lw + 2lh + 2wh = 1704 ---> lw + lh + wh = 852 ***
4l + 4w + 4h = 208 ----> l+w+h = 52 ****
from ****, l = 52-w-h
from ** , l(w+h) + wh = 852 ----> l = (852-wh)/(w+h)
52-w-h = (852-wh)/(w+h)
852-wh = 52w + 52h - w^2-wh -wh - h^2
w^2 + h^2 - 52w + wh - 52h + 852 = 0 obtained only from *** and ****
have not involved ** yet
from ** , l = 4320/(wh)
combine that with l = 52-w-h to get
4320/(wh) = 52-w-h
arghhh, getting too messy to type.
I will give you a hint: w = 18
from 4320/(wh) = 52-w-h, you can now get h
then go back to lwh = 4320 to find l