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Question: ACME Containers ACME Containers Inc. is trying to find the most efficient way of making a new container. The container must have a surface area no larger than 600 square centimeters. Of course, the best container to make is a sphere, but due to the limitations of their manufacturing equipment, they are unable to make a spherical container. The only shapes that they can make are cubes, right cylinders, or right cones. Note that the calculation of surface area must include any tops and bottoms. What container (and with what dimensions) should they make to maximize the volume in the container? Answer: We have three container types, a cube, and right circular cylinder and a right cone. For this we can deduce which container would be the most volume for surface area by analysing their properties, but we would still have to solve for the dimensions. Thus the answer to this problem comes by computation. So let's start with the cube: It is well known that the optimum surface area to volume ratio of a box is found when the box is a regular cube. Simply by taking the closest approximation to a sphere from a box do we get this ratio, and since the optimum surface was given as a sphere, the reader should take this optimum box as a fact. Since the cube has six equal square sides, the area of each is 600/6 100. Since the area of a square is given by two equal sides multiplied together, the length of each side is sqrt(100) 10. Thus to find the volume of the cube we take the area of one side and multiply it by the height of the cube, 100 * 10 1000. Now we consider the right circular cylinder. We know that the circle is a better perimeter to area ratio than the square, much in the same way that the sphere is a better surface area to volume ratio to the cube. Looking at the right circular cylinder as a cube with a circle as a base instead of a square seems to imply that the right circular cylinder is a better surface area to volume ratio than the cube. Using calculus we will show that this is true. We know that the volume of a right circular cylinder is given by: V = Pi*r^2*h where r is the radius of the base and h is the height. The surface area is given by: S = 2*pi*r*h + 2*Pi*r^2, and in this case S = 600. Thus we can solve for h: h = 300/(Pi*r) - r And the volume can be put in terms of r only: V = Pi*r^2*(300/(Pi*r) - r) = r*300 - Pi*r^3 Now we need to find the r where the volume is no longer increasing, nor decreasing, i.e. V' = 0 dV/dr is given by: V'= 300 - 3*Pi*r^2 = 0 Solving for r, we get: r = sqrt(100/Pi) =3D 10*sqrt(1/Pi) We can now solve for h by substituting this value for r into the surface area equation: h = 200/(Pi*10*sqrt(1/Pi)) And plugging these two values into the equation for volume: V = Pi*100/Pi*20/(Pi*sqrt(1/Pi)) = 1128.38 Which is greater than the volume of the box, as we predicted. Now all that remains is the right cone. Using similar methods as above, thought the equations are far more messy, we can solve for the optimum volume of a right cone. V = 1/3*Pi*r^2*h S = Pi*r^2+Pi*r*sqrt(r^2+h^2) =3D 600 h = sqrt(((600 - Pi*r^2)/(Pi*r))^2 - r^2) Thus: V = 1/3*Pi*r^2*sqrt((600-Pi*r^2)^2/(Pi^2*r^2)-r^2) V' = 2/3*Pi*r*sqrt((600-Pi*r^2)^2/(Pi^2*r^2)-r^2)+1/6*Pi*r^2*(-4*(600-Pi*r^2)/ (Pi*r)-2*(600-Pi*r^2)^2/(Pi^2*r^3)-2*r)/(sqrt((600-Pi*r^2)^2/(Pi^2*r^2)-r ^2)) 0 Solving that for r: r 5*sqrt(6)/sqrt(Pi) And finally solving for V: V 1000*sqrt(3)/sqrt(Pi) ~= 977.21 Thus the optimum surface area to volume surface of our three is the = right circular cylinder, with dimemsions: r = 10*sqrt(1/Pi)cm h = 20/(Pi*sqrt(1/Pi)) -- Stefan Dorsett |
