Add Math Sba
By Jabari1
Apr 20, 2015
501 Words
Innovating new packaging by determining the best suitable level of efficiency with the use of calculus for Trini Chocolate Delights.
JABARI BOWMAN
ADD MATH
SBA
CLASS: 56
SCHOOL: ST. MARY’S COLLEGE
TEACHER: MR. YIPYOUNG
Problem Statement
The project’s purpose is to work out the best level of production to suit the new packaging for ‘Trini Chocolate Delights ltd.’ The company’s objective is to maximize produce of their new product but minimize manufacturing cost. To fulfill this task the use of calculus is needed along with other mathematical methods to help design a best suitable and cheap packaging that will be used to carry a grand amount of smaller products while cost remain in a stead/safe amount.
Mathematical Formulation
Below is a list of formulas applied to each question
A. Tan =

B. Substitution
Area of triangle + Area of rectangle= Area of pentagon
C. Differentiation
D. The quadratic equation
The quadratic formula
OR
Factorization of the quadratic formula
E.
Problem solution
A Problem diagram
E B
ycm
D 8xcm C
i) F is the midpoint of line EB and it’s also the perpendicular of triangle EAB. The line AF cuts triangle EAF in half resulting in two isosceles triangles. Since AFB= AFE and DC=EB, then FB and EF are both equal to 4xcm. , , then
Therefore
ii) Since the total crosssectional area is 90cm then:
(i)
(ii)
Make y the subject in equation (i)
Now sub (i) into (ii)
Maximum area
When x=3 the maximum area:
Therefore the maximum crosssectional area is 540
When the equation is divided by 60 it becomes:
Therefore
These are the two possible values of x that satisfies the maximum crosssectional area being 300
X Value
0
1
2
3
4
5
6
0
300
480
540
480
300
0
Prism length=5cm
x=1
Since x=1, crosssectional area=300 and crosssectional perimeter=90cm Volume of prism=
=
When the sides of the container is opened up it makes a net consisting of 7 faces, 2 pentagonal faces and 5 rectangular faces.
Length=Crosssectional perimeter=90cm
Width=prism length=5cm
Therefore
Area of rectangles=590
=450
Total surface area (sum of the areas of each face) = Crosssectional area+ area of rectangles =300(2) +450
=1050
Fixed costs=$6000
Variable costs= and
Average costs (AC) =
For average cost to be a minimum
X=600{Therefore 600 units has to be produced to maintain average cost at its minimum level.}
Since x=600 gives a minimum value
Application of Solution
Since x (the number of units produced) is equaled to 600, the average cost of the product using the formula:
Therefore the possible minimum average cost is $25 per unit
Conclusion
It was determined that the Trini Chocolate delights ltd. had to produce 600 units at the minimum average cost of $25 per tin of chocolates. Also the maximum crosssectional area was determined to be 300 if the value of x was either 5cm or 1cm.
JABARI BOWMAN
ADD MATH
SBA
CLASS: 56
SCHOOL: ST. MARY’S COLLEGE
TEACHER: MR. YIPYOUNG
Problem Statement
The project’s purpose is to work out the best level of production to suit the new packaging for ‘Trini Chocolate Delights ltd.’ The company’s objective is to maximize produce of their new product but minimize manufacturing cost. To fulfill this task the use of calculus is needed along with other mathematical methods to help design a best suitable and cheap packaging that will be used to carry a grand amount of smaller products while cost remain in a stead/safe amount.
Mathematical Formulation
Below is a list of formulas applied to each question
A. Tan =

B. Substitution
Area of triangle + Area of rectangle= Area of pentagon
C. Differentiation
D. The quadratic equation
The quadratic formula
OR
Factorization of the quadratic formula
E.
Problem solution
A Problem diagram
E B
ycm
D 8xcm C
i) F is the midpoint of line EB and it’s also the perpendicular of triangle EAB. The line AF cuts triangle EAF in half resulting in two isosceles triangles. Since AFB= AFE and DC=EB, then FB and EF are both equal to 4xcm. , , then
Therefore
ii) Since the total crosssectional area is 90cm then:
(i)
(ii)
Make y the subject in equation (i)
Now sub (i) into (ii)
Maximum area
When x=3 the maximum area:
Therefore the maximum crosssectional area is 540
When the equation is divided by 60 it becomes:
Therefore
These are the two possible values of x that satisfies the maximum crosssectional area being 300
X Value
0
1
2
3
4
5
6
0
300
480
540
480
300
0
Prism length=5cm
x=1
Since x=1, crosssectional area=300 and crosssectional perimeter=90cm Volume of prism=
=
When the sides of the container is opened up it makes a net consisting of 7 faces, 2 pentagonal faces and 5 rectangular faces.
Length=Crosssectional perimeter=90cm
Width=prism length=5cm
Therefore
Area of rectangles=590
=450
Total surface area (sum of the areas of each face) = Crosssectional area+ area of rectangles =300(2) +450
=1050
Fixed costs=$6000
Variable costs= and
Average costs (AC) =
For average cost to be a minimum
X=600{Therefore 600 units has to be produced to maintain average cost at its minimum level.}
Since x=600 gives a minimum value
Application of Solution
Since x (the number of units produced) is equaled to 600, the average cost of the product using the formula:
Therefore the possible minimum average cost is $25 per unit
Conclusion
It was determined that the Trini Chocolate delights ltd. had to produce 600 units at the minimum average cost of $25 per tin of chocolates. Also the maximum crosssectional area was determined to be 300 if the value of x was either 5cm or 1cm.