The rectangular prism has the greatest surface area
Step-by-step explanation:
Verify the surface area of each container
case A) A cone
The surface area of a cone is equal to
![SA=\pi r^{2} +\pi rl](/tpl/images/0436/0908/68f8d.png)
we have
----> the radius is half the diameter
![l=10\ in](/tpl/images/0436/0908/05522.png)
substitute the values
![SA=(3.14)(3)^{2} +(3.14)(3)(10)=122.46\ in^{2}](/tpl/images/0436/0908/3d00a.png)
case B) A cylinder
The surface area of a cylinder is equal to
![SA=2\pi r^{2} +2\pi rh](/tpl/images/0436/0908/6b0fa.png)
we have
----> the radius is half the diameter
![h=10\ in](/tpl/images/0436/0908/1617d.png)
substitute the values
![SA=2(3.14)(3)^{2} +2(3.14)(3)(10)=244.92\ in^{2}](/tpl/images/0436/0908/a614a.png)
case C) A square pyramid
The surface area of a square pyramid is equal to
![SA=b^{2} +4[\frac{1}{2}bh]](/tpl/images/0436/0908/e67bd.png)
we have
----> the length side of the square
----> the height of the triangular face
substitute the values
![SA=6^{2} +4[\frac{1}{2}(6)(10)]=156\ in^{2}](/tpl/images/0436/0908/eb68e.png)
case D) A rectangular prism
The surface area of a rectangular prism is equal to
![SA=2b^{2} +4[bh]](/tpl/images/0436/0908/404b3.png)
we have
----> the length side of the square base
----> the height of the rectangular face
substitute the values
![SA=2(6)^{2} +4[(6)(10)]=312\ in^{2}](/tpl/images/0436/0908/c5cdb.png)