Question

Awaste management company is designing a rectangular construction dumpster that will be twice as long as it is wide and must hold 19y * d ^ 3 of debris. Find the dimensions of the dumpster that will minimize its surface area.

Answer 1

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:

$S=2(w*h+w*2w+2wh)=2(3wh+2w^2)$

The volume restriction is:

$V=w*2w*h=2w^2h=19\\\\h=\frac{9.5}{w^2}$

replacing h in the surface equation, we have:

$S=2(3wh+2w^2)=6w(\frac{9.5}{w^2})+4w^2=57w^{-1}+4w^2$

Derivate the above equation and set it to zero

$dS/dw=57(-1)w^{-2} + 8w=0\\\\57w^{-2}=8w\\\\w^3=57/8=7.125\\\\w=\sqrt[3]{7.124} =1.92$

The height will be:

$h=9.5/w^2=9.5/(1.92^2)=9.5/2.69=2.58$

Therefore,The dimensions that minimize the surface are:

Wide: 1.92 yd  

Long: 3.83 yd

Height: 2.58 yd