2. Optimizing a Pyramid

3. ...make a pyramid with a square base of maximum possible volume out of a shoebox with the following dimensions: Length=11.5 inches Width= 7.25 inches Height=4 inches MY IS to...

4. The box’s volume is : v=lwh, its surface area is: s=2lw+2wh+2lh l=11.5 in. h= 4 in w=7.25in h b b The pyramid’s volume is: V=b²h/3, this is my primary equation It’s surface area is: S=2bh+b², this is my secondary equation. h=? b=? 11.5” 4” 7.25”

5. he box has a volume of : V= (11.5)(4)(7.25)=333.5 in³ and a surface area of: S=2(11.5)(7.25)+2(7.25)(4)+2(11.5)(4)=316.75 in² I am trying to find the dimensions for the greatest possible volume, but I also know that the surface area of the pyramid cannot be greater than that of the box. It is possible that it is equal to or less than it, so I set the surface area of the pyramid equal to that of the box and I solved for h: 316.75=2bh+b² (316.75-b²)/2b=h

6. ow that I have solved for h, I can substitute it into the equation V=b²h/3 so that: V=(316.75b-b³/2)(1/3)=316.75b-b³/6 and find the derivative: V’=[6(316.75-3b²)+316.75b-b³(0)]/36 =(316.75-3b²)/6 To find the maximum, I set it equal to zero and solve for b: 0=(316.75-3b²)/6 3b²=316.75 b=10.27 in. Then I can plug in b into the formula (316.75-b²)/2b=h to find the value of h: h=[316.75-10.27²]/2(10.27)=10.28 in.

7. I have the values of h and b,now I can find the length of the side of each face of the pyramid: I can take half of a triangle so that the measurements are: 10.28 in. 5.13 in. “c” in. Using the Pythagorean theory I can solve for c: 10.28²+5.13²=c² c=11.5 in.

8. 11.5” 10.27” h=10.28” The surface area of the pyramid is: S=2bh+b² S=2(10.27)(10.28)+10.27² =316.62 in² The volume of the pyramid is : V=b²h/3 =361.63in³. THE END