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Method:
1. Three squares measuring 50 x 50 cm were cut from bristol board sheets using rulers, set squares, pencils and scissors.
It is from this square that the smaller squares of sides (x) will be cut from the edge. 2. The differential of the volume of the box was found, and the value of (x) that would give the maximum volume was found by substituting the (x) values into the second differential. 3. Then smaller squares of size (x), which was found to be 8.33 x 8.33 cm, were cut from the edges of the 50 x 50 cm square.
The cut shape was then folded and taped to provide the box with the maximum volume. 4. A square of sides 2 cm was cut from the edge of the 50 x 50 cm sheet. The flat shape was also folded and taped to produce a box. 5. A square of sides 20 cm was also cut from the edge of the 50 x 50 cm sheet. The flat shape was also folded to produce a box. 6. Appropriate calculations were made to prove that the square 2 cm and the square 20 cm did not produce a box with the maximum volume.
The concept used to solve the problem
To calculate the volume of the cube, length x breadth x height was utilized. This gave a cubic equation. This equation was then differentiated which gave a quadratic formula. dydx was then equated = 0. The quadratic was then solved using the quadratic formula ( x=b±b24ac2a) to obtain two values of (x). These values were then substituted into the second differential (d2ydx2). If the value substituted produced a negative value, then that will be the length of one side of the square to be cut out from the edges of the 50 x 50 cm square to produce the maximum volume of the box. If the value substituted produced a positive value, it is therefore not the length to be used to produce the maximum volume of the box.
The mathematical equations that were utilized.
1. dydx = the derivative of the volume of the box
2. d2ydx2 =the second derivative of the volume of the box 3. x=b±b24ac2a = the quadratic formula
4. X=length of the side of the square to be cut/cm
5. l= length of cube/cm
6. w=width of the cube/cm
7. h=height of the cube/cm
8. V= l × w × h= volume of cube
Calculations V= l x w x h L= 50 – 2x
W= 50 – 2x
H= x
V= (502x) x (502x) x (x)
= 4×3 – 200×2 + 2500x
dydx = 12×2200x +2500
dydx =0
12×2200x +2500=0
Using the quadratic formula
x=b±b24ac2a
x=400±40024250012 24
X=25 or x=8.33
Substitute the values of x into the second differential
d2ydx2 =24x400
When x=25
d2vdx2=24(25)400
=200
The value is positive when x=25, so therefore the volume is not maximum.
When x=8.33
d2vdx2=24(8.33)400
=200
The value is positive when x=8.33, so therefore the value is maximum.
Justification of answer
Substitute the values of x into the volume
V= 4×3 – 200×2 + 2500x
Substitute x=20
=4(20)3200(20)2+2500(20)
=2000
The box made when 20 cm squares were cut from the edges
The graph above shows that were x=20 , the y coordinate is not the maximum on the curve. Therefore the volume is not the maximum.
Substitute x=2
V=4(2)3200(2)2+2500(2)
=4232
The box made when 2 cm squares were cut from the edges.
=4232
The graph above shows that were x=20, the y coordinate is not the maximum on the curve. Therefore the volume is not the maximum.
Substitute x=8.33
V=4(813)3200(813)2+2500(813)
=9259.333
The box made when 8.33 cm squares were cut for the edges.
the above graph shows that were x=8.33, the y coordinate is the maximum on the curve. Therefore the volume is the maximum at x=8.33
The above graph shows the position of all the coordinates. It is clearly see here that when x=8.33, the curve is at a maximum.
Discussion
By taking this mathematical approach, Mr. Lee was able to successfully solve his problem. He could now fit an 8 inch cake in a box made from the same amount of material used to make the 5 inch boxes. This has many advantages. By using the same material, profits are maximized. He can now sell a bigger cake and make more profits than he was making on the 5 inch cake. If he sells his 5 inch cake for $40 and he sells his 8 inch cake for $60, he can increase his profits by 50%. With this increased profits that he is making, he can: expand his business, hire more workers, improve his facilities, open new branches of his franchise and many other things. He can also increase the size of not only his cakes, but anything that he uses to put in the boxes. Also if Mr. Lee exports his goods, he can ship more of his goods than he used to using the smaller box.
This will aid him in receiving more foreign revenue. Also if he is selling a bigger cake, people will get a better buy for their money. This will boost his popularity and more customers will be attracted to shopping at Mr. Lee’s. This will benefit his business greatly. Also Mr. Lee can put items that normally could not have fit in the small box, in his bigger box. This will lead to a greater variety in his products. He can not only make more money from cakes, but from other items. This will eventually lead to an evolution of his company.
He could expand his business from selling pastries to selling food and other items. This use of a bigger box can also help in the preservation of the environment. Because a bigger cake is being put in a bigger box made from the same amount of material. This will result in less materials being used and therefore less pollution. This method may not only be used by Mr. Lee and cakes, but for many other reasons. It can be used by manufacturing companies, packaging companies, industries, local businesses, post offices, and many others. This will ensure that their profits are maximized and they are working at maximum efficiency.
Cite this page
Add maths SBA. (2016, Dec 16). Retrieved from https://studymoose.com/addmathssbaessay