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)](https://tex.z-dn.net/?f=S%3D2%28w%2Ah%2Bw%2A2w%2B2wh%29%3D2%283wh%2B2w%5E2%29)
The volume restriction is:
![V=w*2w*h=2w^2h=12\\\\h=\frac{6}{w^2}](https://tex.z-dn.net/?f=V%3Dw%2A2w%2Ah%3D2w%5E2h%3D12%5C%5C%5C%5Ch%3D%5Cfrac%7B6%7D%7Bw%5E2%7D)
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](https://tex.z-dn.net/?f=S%3D2%283wh%2B2w%5E2%29%3D6w%28%5Cfrac%7B6%7D%7Bw%5E2%7D%29%2B4w%5E2%3D36w%5E%7B-1%7D%2B4w%5E2)
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](https://tex.z-dn.net/?f=dS%2Fdw%3D36%28-1%29w%5E%7B-2%7D%20%2B%208w%3D0%5C%5C%5C%5C36w%5E%7B-2%7D%3D8w%5C%5C%5C%5Cw%5E3%3D36%2F8%3D4.5%5C%5C%5C%5Cw%3D%5Csqrt%5B3%5D%7B4.5%7D%20%3D1.65)
Then, the height h is:
![h=6/w^2=6/(1.65^2)=6/2.7225=2.2](https://tex.z-dn.net/?f=h%3D6%2Fw%5E2%3D6%2F%281.65%5E2%29%3D6%2F2.7225%3D2.2)
The dimensions that minimize the surface are:
Wide: 1.65 yd
Long: 3.30 yd
Height: 2.20 yd