A rectangular package to be sent by a postal service can have a maximum combined

Step 1:
$l+2(w+h)=108$ ... (Equation 1)
Since we are assuming the cross-section is square, we can set $w=h$. Now, let's express the volume of the rectangular package in terms of $l$ and $w$:
$V=l\times w^2$ ... (Equation 2)
Our goal is to maximize the volume $V$. To do this, we can express $l$ in terms of $w$ using Equation 1 and substitute it into Equation 2. Let's solve Equation 1 for $l$:
$l=108-2(w+h)$
$l=108-2(2w)$
$l=108-4w$
Step 2:
Now, substitute this expression for $l$ into Equation 2:
$V=(108-4w)\times w^2$
$V=108w^2-4w^3$
To find the dimensions of the package that maximize the volume, we need to find the critical points of $V$. We can do this by taking the derivative of $V$ with respect to $w$ and setting it to zero:
$\frac{dV}{dw}=216w-12w^2$
Setting $\frac{dV}{dw}=0$ and solving for $w$, we get:
$216w-12w^2=0$
$12w(18-w)=0$
This equation has two solutions: $w=0$ and $w=18$. Since $w$ represents a length, we discard the solution $w=0$.
Step 3:
Now, we can substitute $w=18$ back into Equation 1 to find the corresponding values of $l$ and $h$:
$l=108-4w$
$l=108-4(18)$
$l=108-72$
$l=36$
$h=w=18$
Therefore, the dimensions of the package with the maximum volume that can be sent are $l=36$ inches, $w=18$ inches, and $h=18$ inches.

Result:
18 inches
Solution:
The perimeter of the cross-section is given by the sum of all four sides, which is $4x$. We are told that the maximum combined length and girth is 108 inches, so we can write the equation as:
$4x+2x=108$
Simplifying the equation, we have:
$6x=108$
Dividing both sides by 6, we get:
$x=18$
Therefore, the side length of the square cross-section is 18 inches.
Now, to find the other dimensions of the rectangular package, we know that the length and width are both equal to the side length of the square cross-section. Thus, the dimensions of the package that maximize its volume are:
Length = Width = Height = 18 inches

The perimeter of a cross-section (girth) can be calculated as the sum of the lengths of all four sides. In this case, since the cross-section is square, the girth is given by $4W$ inches.
According to the problem statement, the maximum combined length and girth of the package is 108 inches. Hence, we can write the following equation:
$L+4W=108$
To find the dimensions of the package with the maximum volume, we need to express the volume as a function of a single variable. The volume of a rectangular package is given by the product of its length, width, and height, which can be represented as:
$V=L·W·W^2=L·W^3$
Now, we can substitute the value of $L$ from the previous equation into the volume equation:
$V=(108-4W)·W^3$
To find the maximum volume, we'll take the derivative of the volume function with respect to $W$ and set it equal to zero:
$\frac{dV}{dW}=0$
Differentiating the volume equation with respect to $W$, we get:
$\frac{dV}{dW}=-4W^3+3(108-4W)W^2$
Setting this derivative equal to zero and solving for $W$, we can find the critical points:
$-4W^3+3(108-4W)W^2=0$
Simplifying further:
$-4W^3+324W^2-12W^3=0$
Factoring out $W^2$:
$W^2(-4W+324-12W)=0$
This equation yields two possible solutions: $W=0$ or $-4W+324-12W=0$.
We discard $W=0$ since it doesn't make sense in the context of the problem. Solving $-4W+324-12W=0$ for $W$, we find:
$-16W+324=0$
$16W=324$
$W=\frac{324}{16}$
Simplifying further:
$W=20.25$
Now that we have the value of $W$, we can substitute it back into the equation $L+4W=108$ to find $L$:
$L+4(20.25)=108$
$L+81=108$
$L=27$
Therefore, the dimensions of the package of maximum volume that can be sent are $27$ inches for the length ($L$) and $20.25$ inches for the width and height ($W$).