View Question
 Question
 Subject: optimization problem Category: Miscellaneous Asked by: miry16-ga List Price: \$50.00 Posted: 17 Mar 2006 09:36 PST Expires: 16 Apr 2006 10:36 PDT Question ID: 708442
 ```Your boss wants you to design a container to hold 1000 in3 of popcorn. The container will have a square base and a separate top. The top will come down 2 inches to overlap the rest of the container. Part[1] Find the dimensions of the box which minimize the amount of materials used to construct the box. Part[2] The material used for the top piece costs twice as much per square inch as the material used to construct the other piece of the box. Find the dimensions of the box which minimizes the cost of production. Part[3] Your boss asks you if it would be better to have the shape of the base, something besides a square. Repeat Part[2] with the base of the box several different shapes (at least three.) Find at least one other shape which would lower the cost of the box but still contain 1000 in3.```
 ```HI!! Part[1] Find the dimensions of the box which minimize the amount of materials used to construct the box. Here you must find which dimensions minimize the Surface function: You know that V = x^2*y = 1000 ,then y = 1000/x^2 , where x is the length of the square side and y is the height of the box. S = 2*x^2 + 4*x*y = = 2*x^2 + 4*x*(1000/x^2) = = 2*x^2 + 4000/x S'(x) = 4*x - 4000/x^2 and S''(x)= 4 + 8000/x^3 To find where S(x) has a minimum we need to equal its first derivative to zero: 4*x - 4000/x^2 = 0 <==> 4*x = 4000/x^2 <==> x^3 = 1000 <==> x = 10 . S''(10) = 4 + 8 = 12 > 0, then x = 10 is a minimum. Then y = 1000/x^2 = 1000/100 = 10. The dimensions of the box which minimize the amount of materials used to construct it are 10" for each base side and 10" height (a cube!!). --------------- Part[2] The material used for the top piece costs twice as much per square inch as the material used to construct the other piece of the box. Find the dimensions of the box which minimizes the cost of production. Here we must repeat the procedure but minimizing the Cost function: You know that V = x^2*y = 1000 ,then y = 1000/x^2 , where x is the length of the square side and y is the height of the box. C = x^2 + 4*x*y + 2*x^2 = (the last term shows that the top costs the double than the base) = 3*x^2 + 4*x*1000/x^2 = = 3*x^2 + 4000/x C'(x) = 6*x - 4000/x^2 and C''(x) = 6 + 8000/x^3 To find where C(x) has a minimum we need to equal its first derivative to zero: 6*x - 4000/x^2 = 0 <==> 6*x = 4000/x^2 <==> x = 8.7358 C''(8.7358) = 18 ,then x = 8.7358 is a minimum of the cost. Then y = 1000/x^2 = 13.1037 The dimensions of the box which minimizes the cost of production are 8.7358 inches for each side of the base and 13.1037 inches of height. The minimum cost is: C(8.7358) = = 3*(8.7358)^2 + 4000/8.7358 = 686.8285 ---------------- Part[3] Your boss asks you if it would be better to have the shape of the base, something besides a square. Repeat Part[2] with the base of the box several different shapes (at least three.) Find at least one other shape which would lower the cost of the box but still contain 1000 in3. The shape of the base is a CIRCLE with radius r and the height of the box is h: V = PI*r^2*h = 1000 ==> h = 1000/(PI*r^2) Using the same method of Part 2: C = 3*PI*r^2 + 2*PI*r*h = = 3*PI*r^2 + 2*PI*r*1000/(PI*r^2) = = 3*PI*r^2 + 2000/r C'(r) = 6*PI*r - 2000/r^2 and C''(r) = 6*PI + 4000/r^3 6*PI*r - 2000/r^2 = 0 <==> 6*PI*r = 2000/r^2 <==> r^3 = 1000/(3*PI) Then r = 4.73416 Since C''(4.73416) > 0 it is a minimum. h = 1000/(PI*4.73416^2) = 44.6184 The minimum cost in this case is: C(4.73416) = 3*PI*(4.73416)^2 + 2000/4.73416 = 633.6921 ----------------- The shape of the base is an EQUILATERAL TRIANGLE which each side length is x and the height of the box is h: V = sqrt(3)/4 * x^2 * h = 1000 ,then h = 4000/(sqrt(3)*x^2) C = 3*sqrt(3)/4 * x^2 + 3*x*h = = 3*sqrt(3)/4 * x^2 + 3*x*4000/(sqrt(3)*x^2) = = 3*sqrt(3)/4 * x^2 + 3*4000/(sqrt(3)*x) = = 3*sqrt(3)/4 * x^2 + sqrt(3)*4000/x = C'(x) = 6*sqrt(3)/4 * x - sqrt(3)*4000/x^2 C''(x) = 6*sqrt(3)/4 + sqrt(3)*8000/x^3 6*sqrt(3)/4 * x - sqrt(3)*4000/x^2 = 0 <==> <==> 6*sqrt(3)/4 * x = sqrt(3)*4000/x^2 <==> <==> x^3 = 16000/6 <==> x = 13.8672 Since C''(13.8672) > 0 it is a minimum. Then h = 4000/(sqrt(3)*13.8672^2) = 12.0094 The minimum cost in this case is: C(13.8672) = 3*sqrt(3)/4 * (13.8672)^2 + sqrt(3)*4000/13.8672 = = 749.4149 -------------------- The shape of the base is an HEXAGON which each side length is x and the height of the box is h: V = 3*sqrt(3)/2 * x^2 * h = 1000 ,then h = 2000/(3*sqrt(3)*x^2) C = 3*3*sqrt(3)/2 * x^2 + 6*x*h = = 9*sqrt(3)/2 * x^2 + 6*x*2000/(3*sqrt(3)*x^2) = = 9*sqrt(3)/2 * x^2 + 4000/(sqrt(3)*x) C'(x) = 9*sqrt(3) * x - 4000/(sqrt(3)*x^2) C''(x) = 9*sqrt(3) + 8000/(sqrt(3)*x^3) 9*sqrt(3) * x - 4000/(4*sqrt(3)*x^2) = 0 <==> <==> 9*sqrt(3) * x = 4000/(sqrt(3)*x^2) <==> <==> x^3 = 4000/(9*sqrt(3)*sqrt(3)) <==> x^3 = 4000/27 <==> <==> x = 5.2913 C''(5.2913) > 0 then x = 5.2913 is a minimum. h = 2000/(3*sqrt(3)*5.2913^2) = 13.7473 The minimum cost inthis case is: C = 9*sqrt(3)/2 * 5.2913^2 + 4000/(sqrt(3)*5.2913) = = 654.6742 -------------------- I used the following pages for sources on area, surface and volume formulas: "Area Formulas": http://www.math.com/tables/geometry/areas.htm "Surface Area Formulas": http://www.math.com/tables/geometry/surfareas.htm "Volume Formulas": http://www.math.com/tables/geometry/volumes.htm "Areas and Perimeters of Regular Polygons": http://www.algebralab.org/lessons/lesson.aspx?file=Geometry_AreaPerimeterRegularPolygons.xml Search strategy: area volume formulas hexagon formula I hope this helps you. If you find something unclear or some mistake that I could did please feel free to use the clarification feature before rate this answer. I will be glad to give you further assistance on this if it is necessary. Best regards, livioflores-ga``` Clarification of Answer by livioflores-ga on 18 Mar 2006 22:06 PST ```As you can see the boxes with circular and hexagonal shape for the base lower the cost of the square box.```
 ```Livioflores, You missed something in calculating the surface area of the box. You said: S = 2*x^2 + 4*x*y But the problem states: The top [of the box] will come down 2 inches to overlap the rest of the container. So the surface area should be: S = 2*x^2 + 4*x*y + 4*x*2```
 ```Miry, Let V = Volume S = Surface Area The square base has a side of x The box has a height of y The lid of the box covers the top and overlaps the sides of the container by two inches. (1) Minimize the surface area of the box with a square base that contains a volume of 1000 cubic inches. V = x^2*y = 1000 y = 1000/x^2 S = Base + Sides + Lid = x^2 + 4xy + (x^2 + 4*x*2) = 2x^2 + 8x + 4xy = 2x^2 + 8x + 4x*(1000/x^2) = 2x^2 + 8x + 4000/x (2) The cost per square inch is twice as much for the lid as for the rest of the box. Minimize the cost of the box with a square base that contains a volume of 1000 cubic inches. V = x^2*y = 1000 y = 1000/x^2 Let C = Cost Function C = Base + Sides + 2*Lid = x^2 + 4xy + 2*(x^2 + 4*x*2) = 3x^2 + 16x + 4xy = 3x^2 + 16x + 4x*(1000/x^2) = 3x^2 + 16x + 4000/x (3) Vary the shape of the base of the box from a square but let the box still contain 1000 cubic inches. Try three shapes other than a square that would give a lower cost function. The more sides, the greater an area that can be contained by a given perimeter. The limit of these is the circle. A circular base would give the best cost function. A base in the shape of a hexagon or circle would have a lower cost function than a square base, and a base in the shape of a triangle would have a higher cost function. Since this is homework, I'll let you grind the rest of this out.```