Question

A company plan to manufacture closed rectangular boxes that have a volume of 8m3

Answer 1

Answer:

Base Side Length =1.59m

Height = 3.16 m

Step-by-step explanation:

Volume of the box = $8 m^3$

Let the base dimensions = x and y

Let the height of the box =h

However, for any optimal configuration, Width = Length as varying the length and width to be other than equal reduces the volume for the same total(w+l)

Volume, V=xyh=8

Since x=y

$h=\dfrac{8}{x^2}$

Surface Area of the box

$= 2(x^2+xh+xh)\\=2x^2+4xh$

The material for the top and bottom costs twice as much as the material for the sides.

Let the cost of the sides =$1 per square meter

Cost of the material for the sides = 4xh

Cost of the material for the top and bottom = $=2*2x^2=4x^2$

Therefore:

Total Cost, $C= 4xh+4x^2$

Substitution of  $h=\frac{8}{x^2}$  into C

$C= 4x(\frac{8}{x^2})+4x^2\\=\frac{32}{x}+4x^2\\C(x)=\dfrac{4x^3+32}{x}$

To minimize C(x), we find its derivative and solve for the critical points.

$C'(x)=\dfrac{8x^3-32}{x^2}\\ Setting C'(x) to zero\\8x^3-32=0\\8x^3=32\\x^3=4\\x=\sqrt[3]{4}\\ x=1.59 m$

To verify if it is a minimum, we use the second derivative test

$C''(x)=8+\frac{64}{x^3}\\C''(1.59)=23.92$

Since C''(x) is greater than zero, it is a minimum point.

Recall:

$h=\frac{8}{x^2}\\h=\frac{8}{1.59^2}=3.16 m$

Therefore, the dimensions that minimizes the cost are:

Base Side Lengths of 1.59m; and

Height of 3.16 m