Stupid math question of the day. I have a company that produces two widgets. Widget A costs $16.00 a unit to make and sells for $20. Widget B costs $19 a unit and sells for $23. If you needed to make 2 Widget A’s for every Widget B, and had a fixed administrative cost of $40,000, how many of each units would you need to break even? Damn – I blow at math…
I am not a math wiz but- If you need to make 2 Widget A’s for every Widget B, add the sale price of Widget B and two times the sale price of Widget A (2*20)+23=63 Divide the breakeven point of 40K by the 63 40,000/63= 634.92 or 635 units. Of the 635, 2/3 need to be widget A and 1/3 widget B 635 * 2/3=423 of Widget A and 212 of widget B I could be wrong
On second thought, it should be calculated on profit not on revenue- If you need to make 2 Widget A’s for every Widget B, add the profit of Widget B and two times the profit of Widget A (2*4)+4=12 Divide the breakeven point of 40K by the 12 40,000/12= 3333.333 units. Of the 3333.333, 2/3 need to be widget A and 1/3 widget B 3333.333 * 2/3=2222.222 of Widget A and 1111.111 of widget B Or something like that-
Almost, except that: For profit, read: contribution; and After getting 3333.333 units multiply by the same weights used in calculating the composite contribution of 12, in this case 2 for Widget A and 1 for Widget B. If you use fractional weights that is when you should multiply your composite break-even quantity by a fraction. The final answer therefore: Widget A: 6666.667 units Widget B: 3333.333 units To check this out multiply each unit by its unit contribution, sum the contributions, and subtract fixed costs. You should get a profit of 0. Regards, musasira
Mr. Engineer, Here is the quick answer: You need 6666 pieces of widget A and 3333 pieces of widget B. Here are the algebraic equations: equation (1) is $40,000 + A($16) + B($19) = A ($20) + B($23) also, we know that equation (2) is 1A = 2B Substitute equation (2) into (1) and solve for the variable B (or A if you choose). Variable B comes up to 3333 pieces. Therefore, you know A is 6666. See, Republicans aren't bad mathematicians after all
Javalia5400 is closest but not 100% correct. Item ..............Widget A...... Widget B Production $.........16...............19 Sales $.................20..............23 Profit $...................4...............4 Admin $ 40,000 Breakeven 40,000/(2*4 + 1*4)= 3333.333 whichis the number of Widget B Required and thus 6666.67 of Widget A. Note if you round B off to 3333 and then set A at 2*B you get 6666 (as it is very difficult to make and sell part of a widget) 3333 *4 + 6666*4 = 39,996 which is less then the break even point. Actual product required is 3334 of B and 6668 of A for 40,002 which is the minimum production required to offset the fixed cost, thus is the breakeven point.
Thanks everyone for the assistance. It is greatly appreciated. Randall: Unless you are into hard and fast design (which usually means a semiconductor chip or something totally new), most items are designed by the integration method. This means you are integrating parts, components, and assemblies that are already commercially available. This is more true for the semiconductor business as it is prudent to use items already out there than to redesign the wheel everytime you are involved in a new product. Additionally, I have a library of reference material in my cube. I use Trig quite a bit for vector analysis, but again, don't really pay too much attention to the mundane details as I don't have the time (too few engineers, too little time)
Fractional solution Fractional solutions appear in a number of situations and are sometimes meaningful. Consider the following: If I produce 6 units every 8 hours, how many units do I produce per hour? If a business sells 11 units every fortnight, how many units do they sell per week? In some cases it is possible to produce and sell fractions, for example when output is in liquid form, eg 1.25 litres regards, musasira
Mr. Engineer: Thank you for this post; it illustrates a peculiar truth that I still have a hard time accepting. In my undergrad, I took math through vector calculus and ordinary diffeques. I pulled my first "C" in the latter class and decided I'd never be an engineer. This was foolish of me. I have since learned that I had, in fact, acquired the vast majority of higher math I would have needed AND that engineers almost NEVER use higher math in practice! Most of the problems are already solved, you see. In short, I quit too soon. Oh, well. Live and learn.
Not at all. Your bank balances at the end of the month are probably MUCH higher now, you enjoy much more social prestige, and have the security that your job won't be outsourced ever. But definitely, getting a C in a math course that it is not that easy is not a reason to quit, IMHO. I think a great deal of engineers face on a daily basis questions as silly as the one proposed here by Mr. Engineer.
Muy amable. Es verdad que me gusta el derecho y, tambien, siempre he ganado sufficiente. Pero, si fuera posible, seria mejor a ser ambos!
I get to solve problems like this quite frequently when conducting trade studies. The major difference is that I usually have to also figure in the cost of money.
