Answer:
- Base Length = 7.368cm
- Height = 0.737cm.
Explanation:
Volume of the jewelry box=
![40cm^3](https://img.qammunity.org/2021/formulas/mathematics/middle-school/puqk4irt4e56mpx44iokwuouyqtb9o5bis.png)
The box has a square base and is to be built with silver plated sides and nickel plated top and base.
Therefore: Volume = Square Base Area X Height = l²h
![l^2h=40\\h=(40)/(l^2)](https://img.qammunity.org/2021/formulas/mathematics/college/nsuiiwame07bx4s6uoo2k3u6sen83tu9fe.png)
Total Surface Area of a Cuboid =2(lb+lh+bh)
Since we have a square base
Total Surface Area =
![2(l\²+lh+lh)](https://img.qammunity.org/2021/formulas/mathematics/college/zpm4x52uqej079dm41pyscgl8cgh5o2sfl.png)
The Total Surface Area of the box
![=2l\²+4lh](https://img.qammunity.org/2021/formulas/mathematics/college/x69ek7zeb818oglq2vmfdn4hbx4bnca5l9.png)
Nickel plating costs $1 per
![cm\³](https://img.qammunity.org/2021/formulas/mathematics/college/ppsxmsw5n23k8kpfmh8xnb6gon8u2k9tug.png)
Silver Plating costs $10 per
![cm\³](https://img.qammunity.org/2021/formulas/mathematics/college/ppsxmsw5n23k8kpfmh8xnb6gon8u2k9tug.png)
Since the sides are to be silver plated and the top and bottom nickel plated:
Therefore, Cost of the Material for the jewelry box
![=1(2l\²)+10(4lh)](https://img.qammunity.org/2021/formulas/mathematics/college/sn19n6t6mkqo56jjzlekxejclvl64yxpl7.png)
![Cost, C(l,h)=$(2l\²+40lh)](https://img.qammunity.org/2021/formulas/mathematics/college/jdg07mx5lj6adgagh02rs9owxrat9wa8zc.png)
Recall earlier that we derived:
![h=(40)/(l^2)](https://img.qammunity.org/2021/formulas/mathematics/college/wzp0ffrdfgvpbpwcmpuw97ilaav6gbpv26.png)
Substituting into the formula for the Total Cost
![Cost, C(l)=2l\²+40l((40)/(l^2))\\=2l\²+(1600)/(l)\\C=(2l^3+1600)/(l)](https://img.qammunity.org/2021/formulas/mathematics/college/6nikjpb4pg98tl8m7f3ma3n5yvfy8ec98v.png)
The minimum costs for the material occurs at the point where the derivative equals zero.
![C^(')=(4l^3-1600)/(l^2)](https://img.qammunity.org/2021/formulas/mathematics/college/wsxd3qwu1gmlsj3eahruy5xc8d63v0cgls.png)
![4l^3-1600=0\\4l^3=1600\\l^3=400\\l=\sqrt[3]{400}=7.368 cm](https://img.qammunity.org/2021/formulas/mathematics/college/jd5zxxcxpjxzsr9a5rt8qxpmcp6tydpjov.png)
Recall:
![h=(40)/(l^2)=(40)/(7.368^2)=0.737cm](https://img.qammunity.org/2021/formulas/mathematics/college/qjlqovren5gyhx0uvcluiopu06e1tb9ae7.png)
The box which minimizes the cost of materials has a square base of side length 7.368cm and a height of 0.737cm.