Respuesta :
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)]
[tex]h=\frac {20}{2b^2} \\\\h= \frac {10}{b^2} ...(ii)[/tex]
The material required= surface area, S, of the open box
S= lb+2bh+2lh
By using equation (i) and (ii), we have
[tex]S= 2b(b)+2b\times \frac {10}{b^2} +2(2b)\times \frac {10}{b^2}[/tex]
[tex]S= 2b^2+ \frac {20}{b} + \frac {40}{b} \\\\S= 2b^2+ \frac {60}{b} \cdots(i)[/tex]
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
[tex]\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}[/tex]
b=2.47 feet.
Now, by second differentiation, checking the nature of extremum value.
[tex]\frac {d^2S}{db^2}=4+\frac {120}{b^3}[/tex]
For [tex]b>0, \frac {d^2S}{db^2} >0[/tex]
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.
