Can you tell me what id did wrong I got a similar answer but somehow switched the variables. I am supposed to get that the prism length to the triangleedge is l:s 1:√3 but i got √3:1Can't figure out what I did wrong.
Example Problem
8. A producer of nutrition bars is designing a bar that will just fit inside
a package in the shape of an equilateral triangular-based prism.
a) Determine the ratio of the prism length to the triangle edge of a bar that requires a
minimum amount of packaging material.
solution
The volume of an equilateral triangular-based prism is the area of the base x the length. s is the triangle edge and l is the prism length. It will help if you draw a diagram.
The area of an equilateral triangle is s²√3
4
The volume of the equilateral triangular-based prism would be then be (l)s²√3
4
The surface area of the equilateral triangular-based prism is 2 x equilateral triangle area + 3 x triangle edge x prism length (2)s²√3 +3sl
4
V=(l)s²√3
4
SA=s²√3+3sl
2
The volume of the equilateral triangular-based prism is a constant. Solving the volume equation for l gives l=4V
s²√3
substitution
SA=s²√3+3s(4V)
2 s²√3
SA=s²√3+12V
2 s√3
derivative
SA’=s√3-12(V)
s²√3
set this equal to zero to find the minimum
0=s√3-12(V)
s²√3
s√3=12(V)
s²√3
3s³=12V
V=s³
4
volume equation already solved for l
l=4V
s²√3
substitution
l=4 (s³)
s²√3 (4)
l=s/√3
l:s = √3:1
Example Problem
8. A producer of nutrition bars is designing a bar that will just fit inside
a package in the shape of an equilateral triangular-based prism.
a) Determine the ratio of the prism length to the triangle edge of a bar that requires a
minimum amount of packaging material.
solution
The volume of an equilateral triangular-based prism is the area of the base x the length. s is the triangle edge and l is the prism length. It will help if you draw a diagram.
The area of an equilateral triangle is s²√3
4
The volume of the equilateral triangular-based prism would be then be (l)s²√3
4
The surface area of the equilateral triangular-based prism is 2 x equilateral triangle area + 3 x triangle edge x prism length (2)s²√3 +3sl
4
V=(l)s²√3
4
SA=s²√3+3sl
2
The volume of the equilateral triangular-based prism is a constant. Solving the volume equation for l gives l=4V
s²√3
substitution
SA=s²√3+3s(4V)
2 s²√3
SA=s²√3+12V
2 s√3
derivative
SA’=s√3-12(V)
s²√3
set this equal to zero to find the minimum
0=s√3-12(V)
s²√3
s√3=12(V)
s²√3
3s³=12V
V=s³
4
volume equation already solved for l
l=4V
s²√3
substitution
l=4 (s³)
s²√3 (4)
l=s/√3
l:s = √3:1