Answer:
1.92 yd x 3.83 yd x 2.58 yd
Step-by-step explanation:
We have given a rectangular base, that its twice as long as it is wide.
It must hold 19 yd³ of debris.
Lets minimize the surface area, subject to the restriction of volume (19 yd³)
The surface is given by:
[tex]S=2(w*h+w*2w+2wh)=2(3wh+2w^2)[/tex]
The volume restriction is:
[tex]V=w*2w*h=2w^2h=19\\\\h=\frac{9.5}{w^2}[/tex]
replacing h in the surface equation, we have:
[tex]S=2(3wh+2w^2)=6w(\frac{9.5}{w^2})+4w^2=57w^{-1}+4w^2[/tex]
Derivate the above equation and set it to zero
[tex]dS/dw=57(-1)w^{-2} + 8w=0\\\\57w^{-2}=8w\\\\w^3=57/8=7.125\\\\w=\sqrt[3]{7.124} =1.92[/tex]
The height will be:
[tex]h=9.5/w^2=9.5/(1.92^2)=9.5/2.69=2.58[/tex]
Therefore,The dimensions that minimize the surface are:
Wide: 1.92 yd
Long: 3.83 yd
Height: 2.58 yd
