Question

An open top box is to be built with a rectangular base whose length is twice its width and with a volume of 36 ft 3 . Find the dimensions of the box that minimize the materials used.

Answer 1

Answer:

The dimensions of the box that minimize the materials used is $6\times 3\times 2\ ft$

Step-by-step explanation:

Given : An open top box is to be built with a rectangular base whose length is twice its width and with a volume of 36 ft³.

To find : The dimensions of the box that minimize the materials used ?

Solution :

An open top box is to be built with a rectangular base whose length is twice its width.

Here, width = w

Length = 2w

Height = h

The volume of the box V=36 ft³

i.e. $w\times 2w\times h=36$

$h=\frac{18}{w^2}$

The equation form when top is open,

$f(w)=2w^2+2wh+2(2w)h$

Substitute the value of h,

$f(w)=2w^2+2w(\frac{18}{w^2})+2(2w)(\frac{18}{w^2})$

$f(w)=2w^2+\frac{36}{w}+\frac{72}{w}$

$f(w)=2w^2+\frac{108}{w}$

Derivate w.r.t 'w',

$f'(w)=4w-\frac{108}{w^2}$

For critical point put it to zero,

$4w-\frac{108}{w^2}=0$

$4w=\frac{108}{w^2}$

$w^3=27$

$w^3=3^3$

$w=3$

Derivate the function again w.r.t 'w',

$f''(w)=4+\frac{216}{w^3}$

For w=3, $f''(3)=4+\frac{216}{3^3}=12 >0$

So, it is minimum at w=3.

Now, the dimensions of the box is

Width = 3 ft.

Length = 2(3)= 6 ft

Height = $\frac{18}{3^2}=2\ ft$

Therefore, the dimensions of the box that minimize the materials used is $6\times 3\times 2\ ft$