Thefigure below is an outline of a closed rectangular box. Our problemhere is to design the box in such a way that the cost of the box isminimum. We will use partial derivatives in determining thedimensions of the box in such a way that the dimensions will alsosatisfy certain constraints. The box is to be of volume cubic Vcm3. The constraints to be satisfied are that the cost of material forthe front and back sides is b dollars per square cm c dollars persquare cm for the left and right two sides, and d dollars per squarecm for the top and bottom sides where b, c and d are arbitraryconstants. Let the Lengthof the box be x, the width of the box be y, and the height of the boxbe z as indicated in the figure.
Thevolume of the rectangular box will be given by the expression V =xyz. The cost of the front and back faces is $b per cm3hence the cost of the two faces will be 2bxz. The cost left side andright side faces is $c per cm3hence the cost of the two faces will be 2cyz. The cost of the topand bottom side faces is $d per cm3hence the cost of the two faces will be 2dxy. Let total cost of therectangular box be k.
Anexpression for the total cost of the rectangular box will be given bysumming up the cost of the different faces hence k = 2bxz + 2cyz +2dxy.
Thiswill be our objective function which we wish to optimize to getdimensions for a box of minimal cost which also meets the conditionsspecified. To get the minimum cost of the box, we will find out thefirst and second order partial derivatives for the objective functionwith respect to x, y and z. The critical values for the dimensionswill be given when the derivatives are equated to zero.
Thevolume is a function of area and therefore the cost can be written interms of volume or area as indicated below:
The first order partial derivative of k with respect to x is dk/dx = 2bz + 2dy. This gives the two dimensions below
The first order partial derivative of k with respect to y is
dk/dy= 2cz + 2dx. This gives the two dimensions below
The first order partial derivative of k with respect to z is
dk/dz= 2bx + 2cy. This gives the two dimensions below
Toobtain a minimum value for the dimensions, the 2ndorder derivative is derived from the respective 1storder derivative above. However, all the 2ndorder derivatives give zero indicating that the minimum dimensionshould be at zero. Since costs are positive values and zero will notcarry any meaning, it is the constraint of cost given by k that willdetermine the dimensions that will give a minimum cost.
Fromthe expressions above, we can get the dimensions as follows.
z = dy/b
z = dx/c
y = bx/c
Therefore,there are two possibilities for the dimensions of each side of thebox as shown above. Hence the dimensions of the box that will give aminimum value for k, the total cost of the box will be obtained froma combination of the dimensions derived above that is
x = cz/d or x = cy/b
Andin ration form as x/z = c/d or x/y = c/b
y = bz/d or y = bx/c
Andin ration form as y/z = b/d or y/x = b/c
z = dy/b or z = dx/c
Andin ration form as z/y = d/b or z/x = d/c
Thisimplies that the dimensions should be worked out as ratios betweenone side and the other regardless of the units and the respectivecosts of the faces. Therefore, the cost is an important factor ingetting the dimensions. But the constraints given also have someunderlying conditions that would be necessary for the design of therectangular box. Three possible conditions for the costs areconsidered below.
If the cost of the faces are equal thus b = c = d, then the ratio of the dimensions that will give minimum cost will be x: y: z = 1:1:1. This means that the rectangular box must be designed in the form of a cube for the cost to be minimized. The minimum volume of the rectangular would be 1cm3 at a minimum cost of $2(b + c + d).
If the cost of the faces is b = c = 2d, then the ratio of the dimensions that will give minimum cost will be x: y: z = 1:1:2. This means that the rectangular box must be designed in such a way that the base is a Square and the height is half the length half the width. The minimum volume of the rectangular box would be 2cm3 at a minimum cost of $2(2b + 2c + d).
If the cost of the faces is b = 2c = 3d, then the ratio of the dimensions that will give minimum cost will be x: y: z = 3:6:2. The minimum volume of the rectangular box would be 36cm3 at a minimum cost of $12(b + 2c + 3d). The rectangle should be designed such that the length is twice the width and thrice the height.
Fromthe working and the results obtained in this paper, it is evidentthat calculus is a tool that can be used in making decisions invarious aspects of life. The solution to the design problem has beenworked out by considering the cost as a variable that has otherconstraints attached to it. In this case, the cost of the box is adetermining factor and different costing levels will result indifferent dimensions of the box. Changing the dimensions of therectangular box without considering the proportions of the cost willnot result in the minimum cost that satisfies the constraints. Thethree conditions of costing that were considered in this paper gavedifferent amounts of minimum costs for the rectangular box, but wecannot dismiss the higher value because it is a minimum in relationto the condition that it was worked out from.
dk/dx = 2bz + 2dy. The critical values will be given by
2bz+ 2dy = 0
dk/dy = 2cz + 2dx. The critical values will be given by
2cz+ 2dx = 0
dk/dz = 2bx + 2cy. The critical values will be given by
2bx+ 2cy = 0
x = -cz/d = -cy/b
y = -bz/d = -bx/c
z = -dy/b = -dx/c
a= b = c
x/z= c/d = 1/1.
x/y= c/b = 1/1
z/y= b/d = 1/1
Thereforex:y:z = 1:1:1.
If the cost of the faces is b = c = 2d then
x/z= c/d d = ½ c
x/z= 2c/c = 2/1
x/y= c/b b = c
x/y= b/b = 1/1
y/z= b/d b = 2d
y/z= 2d/d = 2/1
If the cost of the faces is b = 2c = 3d then
x/z= c/d 2c = 3d
x/z= 3d/2d = 3/2
x/y= c/b b = 2c
x/y= c/2c = ½
y/z= b/d b = 3d
y/z= 3d/d = 3/1