Amber Dunning

MAT267

Brewer 12:40

December 8^th, 2007

Honors Footnote: Designing A Dumpster

            For this project, a local dumpster was located and measured for its dimensions.  The goal was to determine the most cost efficient design for a new dumpster with the same volume.  This is to be achieved by utilizing Chapter 11.7 of the Essential Calculus text book by James Stewart on identifying maximum and minimum values.  By creating a cost equation and taking the partial derivatives f_x and f_y, the critical points can be determined by setting each equation equal to 0.  Once the critical points are determined, these points can be plugged into the equation D = f_xx* f_yy - f_xy².  If D < 0, then this point is a saddle point, if D > 0 and f_xx > 0 then this point is a local minimum, if D > 0 and f_xx < 0 then the point is a local maximum. 

            First, the volume of the original dumpster was found by the equations:

                             (See diagram of dumpster at the end of this project for areas)

Area(i) = 9*9*.5 = 40.5

Area(ii) = 9*63 = 567

Area(iii) = 72*43.5 = 3132

Area(iv) = 14.5*29.5 = 427.75

Area(v) = .5*42.5*34.9 = 71.6

Area(vi) = 7.7*29.5*.5 = 113.58

Sum of areas = 5022.43

Volume = 5022.43*72 = 361,615 in³ or 209.27 ft­³

It was given that the sides, back and front are to be made from 12-guage steel which costs $.70 per square foot; the base is to be made from 10-guage steel which costs $.90 per square foot; a lid, no matter the dimensions will cost $50; and the welding will cost $.18 per foot for material and labor combined.  From these figures the cost equation of a rectangular dumpster was determined to be:

Total Cost = 1.4xz + 1.4yz + .9xy + .32x + .32y + .64z + 50

Where the dimensions are based on a 3-dimentional graph, x=Width, y=Length, and z=Height.  It is also known that the dumpster volume will have to equal 209ft³.  This means

To get the equation into only 2 variables this z equation was plugged into the total cost equation.  This produced the equation:

Total Cost = 1.4(209/y) + 1.4(209/x)+ .9xy + .32x + .32y + .64(209/(xy))+ 50

The partial derivatives of this equation are then taken:

                                                 f_x = -292.6/(x²) + .9y + .32 -133.76/(yx²)

These were then set equal to 0 and solved to get the critical points. First the equations were simplified to:

f_x:     0= -292.6y + .9x²y² + .32x²y - 133.76

f_y:    0=  -292.6x + .9y²x² + .32y²x - 133.76

Multiplying both sides by 10 and subtracting gives:

0 = -2926y + 9x²y² + 3.2x²y -1337.6

-(0 =  -2926x + 9y²x² + 3.2y²x -1337.6 )

0 = -2926y + 2926x +3.2x²y - 3.2y²x

Then simplifying this equation to get the x and y values which will make the equation equal zero:

0 = 2926(x-y)+3.2xy(x – y)

0 = (2926 + 3.2xy)(x-y)

2926+3.2xy = 0    or   x - y = 0    => (x=y)

The first solution, x = -2926/(3.2y) is not feasible since both x and y must realistically be greater than 0.  So x must equal y. 

In order to find the best solution for x and y, I had to use a reasonable value for z, the height.  I chose z = 6 ft.  To get x and y, 6 was plugged into the equation z = 209/(xy).  This resulted in x = y = 5.901.  These numbers were plugged into the cost equation to get the final total cost, $188.12.

          An important consideration in this investigation is that a rectangular designed was used to simplify the minimizing equation.  However, this is not necessarily the best method of design for a dumpster.  For this reason, a construction company might not want to use this particular model because the original design which literally “cuts corners”, would actually be much more cost efficient.