Add Maths Sba

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Ruel Wallace
Add Maths S.B.A
Mr. Teesdale
5-8

Table of Contents
Cover page……………………………………………………………………………………………1 Table of contents…………………………………………………………………………………..2 Title……………………………………………………………………………………………………...3 Problem Statement………………………………………………………………………………..4 Mathematical Formulation…………………………………………………………………….5 Problem Solution…………………………………………………………………………………..7 Application of Solution…………………………………………………………………………14 Conclusion…………………………………………………………………………………………...15

Title
A mathematical investigation of an experimental package design by Trini Chocolate Designs Ltd. with the goal of minimizing volume of content of packaging and manufacturing cost using Calculus, Trigonometry and Pythagoras Theorem.

Problem Statement
To determine how effective a container is, in adequately storing chocolate and how innovative the use of the package will be. Trigonometry Pythagoras Theorem and Calculus is used to determine: 1) The max cross sectional area of the pentagonal prism

2) The minimum value of the contents
3) Amount material needed for packaging
4) The minimum number of containers in order to make a profit

Mathematical Formulation
In part, (a) we insert a perpendicular bisector at A dropping at the midpoint, M; of EB, which results in two right angle triangles. The base of one triangle must be half of the base, DC, which is 8xcm thus making MB 4xcm. Tan ABE is given as 3/4c. Using tanθ = opposite ÷ adjacent A

E B

D C

Now that we have two sides of a right angle triangle, the third side can be calculated using Pythagoras Theorem, which must result in 5xcm In part, (b) to obtain the perimeter of the cross section all outer sides must be added. This length must equal 90cm. The equation must contain x and y. The area can be calculated by adding the area of the rectangle and triangle in the cross section. When adding substitute the x and y terms found earlier. The result should be the area given in the question. In part (c), we first find the area at the maximum point, which is zero. This needs the formula dA/dX so that when differentiated the answer equals 0. The second differential was found to determine the nature of turning point for the maximum and minimum point. d2A/dx2 < 0 maximum

d2A/dx2 > 0 minimum
After transposing we can determine x. The value for x is then substituted in the equation for the area and the area is obtained In part, (d) varying examples of x need to be justified to lower the volume. It is implied that x > 0. Five values were chosen and were justified by substituting the values into the equation for the area. The area closest to the one requested was chosen. It must be noted that if 6cm is chosen the prism collapses 360(6cm) – 60 (6cm2) is equal to 0 In part, (e) value of x was given which coincides with my value for x chosen. Using this value and the length of the prism given we can use the area of the cross section gained previously and multiply it by the length stated. Secondly, the surface area can be calculated by using the area of the cross section and multiply it by two since the prism has two face identical to it. Then do the same for the base and rectangular side as well as the two upper rectangles. Add all the areas together to get the surface area In the last part, we use the function f(x) equaling the sum of the overheads, labor cost and cost per unit. Avg cost = (total fixed cost +...
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