Question

Design a rectangular milk carton box of width w, length l, and height h which holds 452 cm3 of milk. The sides of the box cost 3 cent/cm2 and the top and bottom cost 4 cent/cm2. Find the dimensions of the box that minimize the total cost of materials used.

dimensions = ?

(Enter your answer as a comma separated list of lengths.)

Answer 1

Our box has 6 sides. Obviously, front and back sides are the same. So are top and bottom sides. And left and right sides too.

We can calculate areas of each side. And then multiplying it by its cost we can find its total price.

For example, dimentions of the front side are $w$ and $h$ . So area is $S_1=wh$. Its price is $P_1=3wh$ . In this way we can construct function.

$f(w,l,h)=6wh+6lh+8wl$ is a function which shows us the full price of the box.

Obviously, $wlh=452$ .

We should minimize $f$ providing $wlh=452$ .

Let's construct Lagrange function.

$L(w,l,h)=6wh+6lh+8wl+\lambda (wlh-452)$

Derivatives are:

$\frac{\partial L}{\partial w}=6h+8l+\lambda lh$

$\frac{\partial L}{\partial l}=6h+8w+\lambda wh$

$\frac{\partial L}{\partial h}=6w+6l+\lambda wl$

Next step is to solve the system

$\begin {cases} 6h+8l+\lambda lh=0\\ 6h+8w+\lambda wh=0\\ 6w+6l+\lambda wl = 0\\ wlh=452 \end {cases}$ $\iff$ $\begin {cases} h(6+\lambda l)=-8l\\ 6h+8w+\lambda wh=0\\ w(6+\lambda l)=-6l\\ wlh=452 \end {cases}$ $\iff$ $\begin {cases} h=\frac{-8l}{(6+\lambda l)}\\ 6h+8w+\lambda wh=0\\ w=\frac {-6l}{(6+\lambda l)}\\ wlh=452 \end {cases}$ $\iff$ $\begin {cases} h=\frac{-8l}{(6+\lambda l)}\\ 6\frac{-8l}{(6+\lambda l)}+8\frac {-6l}{(6+\lambda l)}+\lambda \frac{-8l}{(6+\lambda l)}\frac {-6l}{(6+\lambda l)}=0\\ w=\frac {-6l}{(6+\lambda l)}\\ \frac {-6l}{(6+\lambda l)}l\frac{-8l}{(6+\lambda l)}=452 \end {cases}$ $\iff$ $\begin {cases} h=\frac{-8l}{(6+\lambda l)}\\ \frac{2l}{(6+\lambda l)}= \frac{\lambda l^2}{(6+\lambda l)^2}\\ w=\frac {-6l}{(6+\lambda l)}\\ \frac {48l^3}{(6+\lambda l)^2}=452 \end {cases}$ $\iff$ $\begin {cases} h=\frac{-8l}{(6+\lambda l)}\\ 2l(6+\lambda l)= \lambda l^2\\ w=\frac {-6l}{(6+\lambda l)}\\ \frac {48l^3}{(6+\lambda l)^2}=452 \end {cases}$ $\iff$ $\begin {cases} h=\frac{-8l}{(6+\lambda l)}\\ \lambda l=-12\\ w=\frac {-6l}{(6+\lambda l)}\\ \frac {48l^3}{(6+\lambda l)^2}=452 \end {cases}$ $\iff$ $\begin {cases} h=\frac {4l}{3}\\ \lambda l=-12\\ w=l\\ \frac {48l^3}{36}=452 \end {cases}$

From the last equation it follows $l^3=339$ and finally, $l=\sqrt[3]{339}$ (other roots are complex)

$\begin {cases} h=\frac {4\sqrt[3]{339}}{3}\\ \lambda = \frac {-12}{\sqrt[3]{339}}\\ w=\sqrt[3]{339}\\ l=\sqrt[3]{339} \end {cases}$