sherman Posted January 22, 2006 Report Share Posted January 22, 2006 Hey guys I've got a question I was hoping I could get help with. I don't even know where to start on this and was hoping you guys could maybe give me some hints? A soup manufacturer wants to sell it's soup in 500mL cans. The metal for the top and bottom costs $1.20/m^2. The metal for the side costs $0.40/m^2. After the circles for the top and bottom are cut out of rectangle, the remaining metal will be scrapped. Find the dimensions of the can that will minimize the cost of the materials. ANS: 2.54m, 24.68m I was thinking something to do with the cylinder volume formula V = pi(r^2)(h) because it gives you that 0.5L number, but after that I really don't know where to go. Any help would be appreciated. Thanks in advance! :lol: Quote Link to comment Share on other sites More sharing options...
TheBigDog Posted January 22, 2006 Report Share Posted January 22, 2006 Don't forget the cost of the scrap. Quote Link to comment Share on other sites More sharing options...
kamil Posted January 23, 2006 Report Share Posted January 23, 2006 deleted Quote Link to comment Share on other sites More sharing options...
kamil Posted January 23, 2006 Report Share Posted January 23, 2006 Don't forget the cost of the scrap. The question is asking for dimensions not for cost. ISnt the sentence 'the rest of the metal will be scrap' irrelevant to the question. IN other words isnt it there just to trick people. Quote Link to comment Share on other sites More sharing options...
Erasmus00 Posted January 23, 2006 Report Share Posted January 23, 2006 The question is asking for dimensions not for cost. ISnt the sentence 'the rest of the metal will be scrap' irrelevant to the question. IN other words isnt it there just to trick people. No, because you still pay for the metal you throw out. And they want the dimensions of the can that minimize the cost. -Will Quote Link to comment Share on other sites More sharing options...
kamil Posted January 23, 2006 Report Share Posted January 23, 2006 Then i dont get the question. I thought the factory buys circles and rectangles(which are bent to make the side of the can). So tell me what the 'remaining metal' means, are two diferent sets of rectangles purchased(one $1.20/m^2 and one $0.40/m^2) and the cheaper one is bent to make the rectangle, and the more expnsive one is cut into a circle. ANd the remaining metal from the cutting out is the remaining scrap?????:lol: Quote Link to comment Share on other sites More sharing options...
Tormod Posted January 23, 2006 Report Share Posted January 23, 2006 You have to purchase more metal than is needed for production, hence the scrap issue. The assignment assumes that you are the manufacturer. It seems to me the point of the exercise is to design a can that best utilizes the dimensions of the purchased metal so as to minimize the amount of scrap. Quote Link to comment Share on other sites More sharing options...
kamil Posted January 23, 2006 Report Share Posted January 23, 2006 O ok so the metal for the top and bottom of the can has to be purchased as a rectangle and then cut into a circle right? And the metal for the side can be purchased as a rectangle and then bent into a circular shape? Im very sorry but I feel the problem wasnt defined accurately(oh well might be only me *sigh*:lol: ) BTW, is the first term in the answer the radius(or diamaeter) and the second term is the height? Quote Link to comment Share on other sites More sharing options...
CraigD Posted January 23, 2006 Report Share Posted January 23, 2006 A soup manufacturer wants to sell it's soup in 500mL cans. The metal for the top and bottom costs $1.20/m^2. The metal for the side costs $0.40/m^2. After the circles for the top and bottom are cut out of rectangle, the remaining metal will be scrapped. Find the dimensions of the can that will minimize the cost of the materials. ANS: 2.54m, 24.68mI was thinking something to do with the cylinder volume formula V = pi(r^2)(h) because it gives you that 0.5L number, but after that I really don't know where to go.This is a differential Calculus problem. The ANSwer given appears slightly incorrect, and in the wrong units – it should be Radius=~2.752 cm, Height=~21.02 cm, giving a dimension of the can of about 5.503 x 21.02 cm. Here’s how you go about solving itRealize that this problem is asking for you to solve for a single variable – either Radius, or Height. You can express either variable in terms of the otherFind the equation for volume of a cylinder. Sherman has already found it: volume=pi(radius^2)(height)Substitute the volume given (500 mL = 500 cm^3 = 0.0005 m^3 – best units appear to be cm^3)Rearrange this equation to give either height in terms of radius, or radius in terms of height.Find the equation for cost of the can. This will be (area of top & bottom, including waste)(cost of top/bottom metal) + (area of sides)(cost of side metal). Hint: the instructions tell you that you must cut each top and bottom out of a rectangle of metal, discarding any unused pieces. So, for radius r, the rectangle used must be 2r by 4rSubstitute the equation found in step 4 for height or radius, whichever you solved step 4 forRealize that this problem is asking you to find the size of a can of minimum or maximum cost. To do this, you want to solve the first derivative of the equation found in step 6 for a value of 0Find the first derivative the equation from step 6Set it equal to 0 and solveSubstitute the solution into the equation from step 3, and solve for the other, dependent variable.This seems to be a homework problem. We’ve a tradition here at Hypography of not giving away homework problems unless the asker is completely stuck after making several attempts to solve it themself, so Sherman, see if you can solve it with the hints above, and post your results, or, if you get stuck, post as much as you can. C1ay 1 Quote Link to comment Share on other sites More sharing options...
TheBigDog Posted January 25, 2006 Report Share Posted January 25, 2006 Lets inject some reality and see how much more complicated it becomes. 1) Your base material comes manufactured in sheets that are 1m square. 2) The side material is available in rolls with widths in 1cm increments and lengths of 2m. Now you are dealing with setup and scrap constraints in both sub-assemblies, and have a REAL real world problem to solve. How many circles can you get out of the 1m sheet? How many cans do you make out of a 2m roll? Do you trim the edges, or use the natural height of the material? Manufacturing engineering is fun! Bill Quote Link to comment Share on other sites More sharing options...
Qfwfq Posted January 25, 2006 Report Share Posted January 25, 2006 Don't worry about the scrap Sherman! :eek_big: I don't quite agree about the scrap issue and, as it seems to be a classroom problem and no details concerning scrap are given, I tend to think those costs per square metre are meant to be taken quite simplistically, at face value. :lol: Just take it as a matter of minimizing the cost by the right ratio of width to hight, for the required volume. You might call this ratio rho or sigma or something and write cost as a function of that parameter; how do you now find the minimum value? You have to purchase more metal than is needed for production, hence the scrap issue. The assignment assumes that you are the manufacturer. It seems to me the point of the exercise is to design a can that best utilizes the dimensions of the purchased metal so as to minimize the amount of scrap. For those who want to squabble about the scrap: In a real case, this type of scrap is almost always recycled in the manufacturing process itself and the use to scrap ratio depends mainly on the shape employed (circle or rectangle). To be that realistic one would have to know the margin on recycling, which isn't given by the exercise and is worse if the scrap has to be sent back to the supplier of the rolls, and how close the circles can be cut if you wanted the (small) dependence of the ratio on diameter. The problem doesn't give the useful width of the roll, also relevant for working out the diameter dependence (a semicircle is no use as a can top or bottom). :naughty: I doubt Sherman's teachers are after that kind of thing! Quote Link to comment Share on other sites More sharing options...
Tormod Posted January 25, 2006 Report Share Posted January 25, 2006 :eek_big: I doubt Sherman's teachers are after that kind of thing! Like you know what he's after! You North-Italian dimwit you!!! :naughty: Quote Link to comment Share on other sites More sharing options...
kamil Posted January 25, 2006 Report Share Posted January 25, 2006 Don't worry about the scrap Sherman! :naughty: I don't quite agree about the scrap issue and, as it seems to be a classroom problem and no details concerning scrap are given Thats what I said!!! Just take it as a matter of minimizing the cost by the right ratio of width to hight, for the required volume. Really:eek2: , because i think thats what i did but i then edited my post because i thought i made a mistake. | | | | | |____________ Lets just pretend the cost of the width is 10/m, and the cost of the heght is 1/m. The area needs to be 10m^2. If we hav w=5m and h=2m then then the cost would be 52. BUt if we minimize it to h=10, w=1 then the cost would be 11. THis is the lowest possible cost. I thought similair logic could be applied to this problem. Though of course its 3D hear but since the shape of the cross section of the cylindar remains the same through every single point in height of the cylindar, then we can use the same idea. Quote Link to comment Share on other sites More sharing options...
Qfwfq Posted January 25, 2006 Report Share Posted January 25, 2006 Certainly the same idea applies Kamil and w=1 is the correct minimization in your rectangle problem. During a break I tried the calculus for Sherman's problem and I must rectify myself: it's a lot handier to use the radius r as a parameter. That gives the following function to minimize: C® = pi b r^2 + 2aV/r where a is the surface cost for the siding, b that for tops and bottoms and V is guess what. In case you can't guess, V is the volume and pi is pi. Quote Link to comment Share on other sites More sharing options...
Erasmus00 Posted January 25, 2006 Report Share Posted January 25, 2006 Don't worry about the scrap Sherman! :naughty: I don't quite agree about the scrap issue and, as it seems to be a classroom problem and no details concerning scrap are given Ummm... After the circles for the top and bottom are cut out of rectangle, the remaining metal will be scrapped. So you deffinately need to include the scrap into your calculations. Its built into the question. -Will Quote Link to comment Share on other sites More sharing options...
Qfwfq Posted January 25, 2006 Report Share Posted January 25, 2006 I hadn't seen that but it sure gives little info to be realistic! :naughty: At that point one must idealize extensively and reckon the circles to be cut edge to edge and from a sheet with no edges! In this way the cost of each circle is the cost of a hexagon, replace one of the pis in the calculus (not both!) with the appropriate coeffecient. If I haven't also goofed this in my hurry it should be twice root three. Quote Link to comment Share on other sites More sharing options...
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