Question

A company wishes to manufacture a box with a volume of 20 cubic feet that is open on top and is twice as long as it is wide. Find the width of the box that can be produced using the minimum amount of material.

Answer 1

Answer:

2.47 feet

Step-by-step explanation:

Let l, b, and h be the length, width, and height of the box.

As the length is twice the width, so l=2b ...(i)

The volume of the box is 20 cubic feet, so

lbh=20

(2b)bh=20  [by using (i)]

$h=\frac {20}{2b^2} \\\\h= \frac {10}{b^2} ...(ii)$

The material required= surface area, S, of the open box

S= lb+2bh+2lh

By using equation (i) and (ii), we have

$S= 2b(b)+2b\times \frac {10}{b^2} +2(2b)\times \frac {10}{b^2}$

$S= 2b^2+ \frac {20}{b} + \frac {40}{b} \\\\S= 2b^2+ \frac {60}{b} \cdots(i)$

Now, we have the surface area, S, as the function of width, b.

Differentiate S with respect to b then equate it to zero to get the extremum values, as

$\frac {dS}{db}=0 \\\\\frac {d}{db}(2b^2 +60/b)=0 \\\\4b-\frac{-60}{b^2}=0 \\\\4b=\frac{60}{b^2} \\\\b^3=\frac{60}{4} \\\\b=(15)^{1/3}$

b=2.47 feet.

Now, by second differentiation, checking the nature of extremum value.

$\frac {d^2S}{db^2}=4+\frac {120}{b^3}$

For $b>0, \frac {d^2S}{db^2} >0$

So, the width of the box,  b=2.47,  is the minima for the area, S(b).

Hence, the width of the box that can be produced using the minimum amount of material is 2.47 feet.