Question

Find the largest rectangular parallelepiped (box) that can be shipped by parcel post (length plus girth$\mathbf{}\mathbf{=}\mathbf{}\mathbf{108}$in).

Short answer

The resultant answer is to$\mathrm{w}=18\mathrm{in},\mathrm{h}=18\mathrm{in},\mathrm{l}=36$

Step-by-step solution

Given data

The largest rectangular parallelepiped (box) that can be shipped by parcel post (length plus girth $=108$in).

Concept of Partial differential equation

A partial differential equation consists of two or more independent variables, an unknown function (depending on those variables), and partial derivatives of the unknown function with respect to the independent variables.

Simplify the expression

From the given information, it is a typical optimization problem with constraint, where the girth (g) plus the length $l$is kept constant and we need to find the proportions that will maximize the volume $v$.

So, here use the method of elimination, then using the partial derivative is given below:

$l+g=l+2(w+h)=108\mathrm{in}$

Where, $\mathrm{w}$and $\mathrm{h}$are the width and height respectively.

localid="1664353391918" $\begin{array}{rcl}l& =& 108-2(w+h)\\ V& =& wlh\\ V& =& w\times (108-2(w+h\left)\right)h\\ V& =& 108wh-2w^{2}h-2w{h}^2\end{array}$ ……. (1)

Now find the extreme points at $\begin{array}{rcl}{V}_w& =& 0\end{array}$and $\begin{array}{rcl}{V}_h& =& 0\end{array}$.

$\begin{array}{rcl}\frac{\partial V}{\partial w}& =& 108h-4wh-2h^2=0\end{array}$ …….(2) $\begin{array}{rcl}\frac{\partial V}{\partial w}& =& 108w-4wh-2w^2=0\end{array}$ ..….. (3)

Subtract the partial derivative above from each other.

$\begin{array}{rcl}108h-108w-2h^2+2w^2& =& 0\\ 54(h-w)-(h-w)(h+w)& =& 0\\ (h-w)[54-(h+w\left)\right]& =& 0h\\ & =& w\end{array}$

Substitute the values

Substitute the value of $w$in equation (2).

role="math" localid="1664356948168" $\begin{array}{rcl}54h-2h^2-h^2& =& 0\\ h(54-3h)& =& 0\\ h& =& w\\ & =& 18\mathrm{in}\\ & & \end{array}$

Now substitute in equation (1).

$\begin{array}{rcl}l& =& 108-4h\\ l& =& 108-4\times 18\\ l& =& 36\mathrm{in}\\ & & \end{array}$

Now to check whether the values calculated for h, w, and $\begin{array}{rcl}& & l\\ & & \end{array}$are indeed maximum points.The second derivative test for partial derivatives as follows:

$\begin{array}{rcl}{V}_{ww}& =& -4h,V_{hh}=-4w,V_{wh}=108-4w\\ V_{ww}& =& -72,V_{hh}=-72,V_{wh}=36\\ f_{xx}& <& 0,f_{yy}<0,f_{xx}{f}_{yy}>{f_{xy}}^2\\ & & \end{array}$

This is the criteria of maximum points.

Thus, the solutions are $\begin{array}{rcl}\mathrm{w}& =& 18\mathrm{in},\mathrm{h}=18\mathrm{in},\mathrm{l}=36\mathrm{in}\\ & & \end{array}$.