Question

A waste management company is designing a rectangular construction dumpster that will be twice as long as it is wide and must hold 12 yd cubed of debris. Find the dimensions of the dumpster that will minimize its surface area.

Answer 1

Answer:

The dimensions that minimize the surface are:

Wide: 1.65 yd

Long: 3.30 yd

Height: 2.20 yd

Step-by-step explanation:

We have a rectangular base, that its twice as long as it is wide.

It must hold 12 yd^3 of debris.

We have to minimize the surface area, subjet to the restriction of volume (12 yd^3).

The surface is equal to:

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

The volume restriction is:

$V=w*2w*h=2w^2h=12\\\\h=\frac{6}{w^2}$

If we replace h in the surface equation, we have:

$S=2(3wh+2w^2)=6w(\frac{6}{w^2})+4w^2=36w^{-1}+4w^2$

To optimize, we derive and equal to zero:

$dS/dw=36(-1)w^{-2} + 8w=0\\\\36w^{-2}=8w\\\\w^3=36/8=4.5\\\\w=\sqrt[3]{4.5} =1.65$

Then, the height h is:

$h=6/w^2=6/(1.65^2)=6/2.7225=2.2$

The dimensions that minimize the surface are:

Wide: 1.65 yd

Long: 3.30 yd

Height: 2.20 yd