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Problem 13-3Let S represent the amount of steel produced (In tons): Steel production related to the amount of labor used (L) and the amount of capital used (C) by t...

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Problem 13-3Let S represent the amount of steel produced (In tons): Steel production related to the amount of labor used (L) and the amount of capital used (C) by the following function: 5 = 20 L0.30 C 0.70In thls formula represents the unlts of labor Input and the unlts of capltal Input: Each unit of labor costs $50, and each unit of capital costs S100,(a) Formulate an optimization problem that will determine how much abor and capital are needed In order t0 produce 50,000 tons of steel at minim

Problem 13-3 Let S represent the amount of steel produced (In tons): Steel production related to the amount of labor used (L) and the amount of capital used (C) by the following function: 5 = 20 L0.30 C 0.70 In thls formula represents the unlts of labor Input and the unlts of capltal Input: Each unit of labor costs $50, and each unit of capital costs S100, (a) Formulate an optimization problem that will determine how much abor and capital are needed In order t0 produce 50,000 tons of steel at minimum cost: Min 0.30 0.70 Select your answer L, € Select your answer (b) Solve the optimlzation problem you formulated In part Hint: When using Excel Solver , start wlth an Inltlal If required round your answers to two decimal places_ and C > 0. L = C = Cost



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Computing the cost of Production The Acme Steel Company is a producer of stainless steel and aluminum containers. On a certain day, the following stainless steel containers were manufactured: 500 with 10 -gallon (gal) capacity, 350 with 5-gal capacity, and 400 with 1-gal capacity. On the same day, the following aluminum containers were manufactured: 700 with 10-gal capacity, 500 with 5-gal capacity, and 850 with 1-gal capacity. (a) Find a 2 by 3 matrix representing these data. Find a 3 by 2 matrix to represent the same data. (b) If the amount of material used in the 10 -gal containers is 15 pounds (lb), the amount used in the 5-gal containers is 8 lb, and the amount used in the 1-gal containers is 3 lb, find a 3 by 1 matrix representing the amount of material used. (c) Multiply the 2 by 3 matrix found in part (a) and the 3 by 1 matrix found in part (b) to get a 2 by 1 matrix showing the day's usage of material. (d) If stainless steel costs Acme $\$ 0.10$ per pound and aluminum costs $\$ 0.05$ per pound, find a 1 by 2 matrix representing cost. (e) Multiply the matrices found in parts (c) and (d) to find the total cost of the day's production.

This following problem gives us the following information. It costs $50 per date per worker. Her day for worker it costs $100 per unit manufactured. And it's asking us to write an equation in terms of X, which are the workers, why the units manufactured and see which is the total cost. So given the information that we have, we're looking for the total cost. Therefore, we're looking for C C equals Harry tells us that the worker variable is X and it's $50 a day. Her worker. So it's 50 times X plus. So is this. It's $100 per unit. It's a manufacturer union and that variable it's y 100 times. Why? So this turns out to be the equation. Then we come up with

I told that this is the weekly cost to produce, what was it? X tons of steel. So X. Is the number of tons of steels we steal we produce in a week and this is how much it costs to do that as a function of how much we produce. And so they asked us for what basically incremental cost Have going for 500 tons to 501 tons. So see it 500 is 29,000, See at 501 is 29,105 1. And so the difference and again the increments is just one here. So the difference is 105.1. That would be in dollars. This is dollars too. So To produce an extra an extra ton of steel, if we've already produced 500 we would it would cost $105. So hopefully, you know, maybe that makes sense. Maybe it doesn't um because it depends on what we're selling it for. If we can't sell it for this much then it doesn't make any sense to produce it. Now if we go up to a 1000 and we look at making one more increment, one more ton of steel. So see it 1000 is $1 $106,500. And C. At 1000 and one is $106,705.10 attend something. I guess I didn't carry out to the sense here. So see it one on 101 minus C. C. At 1000 and one minus C. At 1000 divided by one increments. So one extra ton of steel. Now it's going to cost us $205.10 cents To make that one extra ton of steel. So you know what is saying is that as we make more and more it's costing us more and more. And we can particularly see that because this is an increasing function of X. Right? Um so at some point we're gonna say well it's costing us too much to make more because we can't sell it for that much. So if we can't sell a ton of steel for $205 Then it doesn't make any sense to make that extra ton. And at some point in this in between here somewhere, so say we can sell a ton of steel for $150 at some point in here, We're gonna find between a 500,000 we're gonna find where at some point it's going to basically be not worth us worth it to produce that extra ton of steel because we're going to have to sell it at a loss

Okay. This question gives us the demand function and cost function for a certain product, and then it wants us to find the number of units to produce to maximize our revenue. So what we're gonna be doing is setting up an equation for a function. So we want profit to be equal to revenue. So the money we make minus the cost, which is the money we spend to make the product. And this is related to the functions already given because the revenue would be if we created X units at a price given at p of X from the demand function. Because, remember, if we have a demand function p the revenue is just x times P because it's that many units times p dollars per unit. And then we already are given the cost function, which in this case is a of X. So I'm just going to call this function and of X for net. So we get X times p of X, and we said P of X was going to be equal to 2 56 minus 50 x, and then we're going to subtract the cost. And we said the cost function was 1 82 plus 56 x. Okay, Yeah. And now we're just going to start taking this function here and expanding it into a form that's easier to work with. So an FX will be equal to 2. 56 X minus 50 X squared minus 1 80 to minus 56 X. Then we're just gonna spend some time to combine like terms here, and we're going to get kids of minus 50 X square and then to 56 minus 56 is just 200. So plus 200 X and then minus 1 82. And now that we have our function, we're going to try and maximize it. And remember, we maximize functions by setting the derivative equal to zero. So we want to find end prime of X, and I know it's going to be a maximum because I have a downward opening proble so it will look something like this. We're trying to find this maximum right there, so and prime would be negative 100 x plus 200. Or if I set that equal to zero here we get X equals two is our maximize er, so we need to produce two units to maximize our profit and to see what that profit would end up being. We just plug in to to our net equation, which would be two times p of two minus a of to or alternatively, we could just plug into the form up here, which would be okay. Negative 50 times two squared, plus 200 times two and then minus 1 82. And that would just be while negative 50 times four is negative 200 plus 400 minus 1, 82 or 200 minus 1 82 which would be 18 as our profit.

Yes. Um A problem where we have. Let's see here. The production function for particular products. So this is the production function basically kind of. I just call this the number of units. Um This is the units of things reproduced access the number of units of labor. And why is the number of units of capital? Yeah. And so one unit of labour costs $40. And each unit of capital, one unit of capital costs $80. And we have $400,000 to budget. So um are constraint? Is that 40 X plus 80 Y. Has to equal 400,000. So we couldn't write it like this. So we have our lagrange Yeah, function here with our lagrange multiplier. And we can take the partial with respect to X. Y. And lambda. And we do that. And we get this function. These three functions here. So they're not linear obviously. But you know, you can see we have ratios here that makes things a little nice. And you can again just sub packed substitute and stuff. And what you'll find is you get a nice solution, you get X equals 8000. And why it was 1000 Which are plugging into here. It means that the number of units at least rounded off to the thousands is 264,000 units. Now I mentioned in previous on many previous problems and I'm not sure, I wasn't sure exactly what um if the love brash multiplayer will tell you anything and in this kind of contrived example, and they talk about it in the chapter a little bit, it's the whether they call it's the marginal marginal productivity of money, what's the negative of that and why that is it is really, again, in technical, technically in Lagrange multipliers, you could have anything here, you can put two lambda here, you can put lambda square here, you can put land over to here. And so it doesn't really matter what you put here. So it's kind of an arbitrary thing. But if you put it, if this is dollars here, if this is something in dollars or thousands of dollars or whatever, and then this basically gives you the marginal with respect to dollars and it only works really number one, if you know you have something that's a linear function like this. So you have some, you know, some number of dollars per something plus some number of dollars per something else equals sum total number of dollars. And so if that is the case and they don't really go into this a lot of detail, you can show that there is some meaning to this. And what that means is that like, so if you added, so what they ask us to do is add $50,000 in capital. So make this 450 thousands. Well that's gonna do is make this 450,000. Yeah. And so you can see that basically, if you add 450,000, you're gonna you're gonna wind up increasing this. Uh if you add 50,000 a dish, you want to wind up increasing this by this times 50,000 AR- This time 50,000. So it works out that it's kind of but it's again it's not really a general thing that you can say this is the marginal productivity of money. That's only in the case where you have this constraint on money that looks like this, that is some linear function of the of the two things that you're buying. Which is usually the case, I would say, you know, usually, you know, but you know, if you get discounts, say say, you know, the rate goes down as you buy more, then all of a sudden this is not gonna work. I don't think I have to work through the math, but I don't think it will work only in this case. So if we do that, we see that we get Um now we get a number 2,297,000 Units of something instead of this. So we had at 50,000 capital, we get more to re expect. And in fact, if you actually just Make this 50,000 You know, at this, make this 400,000, you get the exact same number just because again, this is all in here. Um Yeah, it's kind of a contrived problem and you can show that, you know, given a constraint like this, then this will scale with minus land at times the scale of this


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