Question 2. Company XYZ produces and sells widgets....

Question 2. Company XYZ produces and sells widgets. Production is done at two separate factories, one in Mississauga and one in Scarborough. An analysis reveals that in the Mississauga location, it costs [ C_{M}(q)=frac{1}{50} q^{3}+10000( ext { dollars) } ] to produce ( q ) widgets in a month, while in the Scarborough location it costs [ C_{S}(q)=q^{2}+20000 ext { (dollars) } ] to produce ( q ) widgets in a month. The company needs to fulfill orders for a total of 400 widgets in the month of March 2023. (a) How many widgets should each location be assigned to produce so that the total cost of production is minimized? (b) Due to differences in quality control and shipping costs, company XYZ decides to sell widgets produced at each factory at different price levels: widgets produced in Mississauga are to be sold for ( $ 2000 ) each, and widgets produced in Scarborough for ( $ 2800 ) each. Show that it is most profitable for the company to allocate all production to Scarborough.

Transcript

00:01 So this problem is how there are, you have an amount of total 400 quantity widgets that you need to be produced.

00:09 And the cost at location m and the cost of location s.

00:13 Part a is how many that you're going to produce at each location to minimize the total cost, right? so for convenient, let's just name q1.

00:24 I don't know.

00:25 So q1 or just q1 to be at the location.

00:30 M or let's name a qm and qs to be at location s the quantity how much you're gonna produce at that and so you always have one equation is qm plus qs is a total amount that you're gonna produce is 400 right and then the cost will be you want to minimize the cost so i'm gonna add both of them cm qm plus cm qs and let's call it to see just the cost what is that going to be not going to be a hundred and fifty of qm q plus qs square plus 3 ,000 i think you can add them right so now i want to minimize this one so this is a function of two variables sorry so how would we minimize this one given a constraint qm plus qm equals equal 400th so that there's a way of doing the gradient of c if you remember it's going to be c of qm partial derivative c of which is expected to both of these okay so let's calculate it that so with respect it to qm that's gonna be 3 over 50 qm square and with respect to qs are going to be 2 qs okay and then this is the constraint let's name it g so you can calculate a constraint of g.

02:05 That's going to be just one in one.

02:08 Okay.

02:09 And then you set them equal.

02:10 So the idea is gradient of c will equal to a constraint lambda gradient of g.

02:17 I don't know this is a laurentian method.

02:19 If you have not done a laurentian method, you can just substitute and make it one variable.

02:25 I don't know what have you learned and what have you not learned, but i prefer to do it this way.

02:31 You can just substitute so let me just like qm equal to 400 minus qs right and then if you substitute into this function you're gonna just you can get one variable okay maybe i should do that well i'm in the middle of this is the lorraine's method there's many method to solve you know what let's let's let's let's try the laran's method okay and then i'm compared the other one if we have time so you're going to say 3 over 50 qm square is lambda time 1 which is lambda and 2 qs is also lambda so therefore because they both lambda so 3 over 50 qm square will equal to qs all right so divide by 2 so 3 over 100 qm square will be qs and then you put into the constraint is that qm plus the qs is 400 so that's qm plus 3 over a hundreds qm square is 400 okay multiply by a hundred so hundreds qm plus three qm square will be 400 and then two more zero and then you're gonna need so um let's write it's in quadratic equation my plus 100 qm minus 4 40 times okay equal zero how do you solve a qm so qm i'm gonna be negative 100 using a quarter over six plus a minus square root of 100 square minus four time three times negative 40 okay and of course you're need a calculator so what inside of the that is plus 490 ,000 and then i'm going to take the square root okay so i got negative 100 plus a minus 700 over 6 so we got two answer but you don't take the negative answer this would be negative 800 over 6 so nope because we don't have negative quantity so let's just take the plus so that would be 600 over 6 and that is exactly 100 so my qm is 100 therefore my qs then be 300 because remember the sum is exactly 400 that will be how you minimize the cost all right so that's part a and that's you i'm using an arrines method.

05:44 Okay.

05:45 So now part b say you are going to have introducing profits.

05:50 You're going to have 2 ,000 if you saw at qm and 2 ,800 that you saw with that.

05:57 So the profit is how 2 ,000 times qm okay minus a c at m plus 2 ,800 how many you saw at qs and minus a c of s.

06:14 And you want to maximize this profit function.

06:16 It's called a p.

06:18 Okay.

06:19 Again, you also have the qm plus the qs is 400.

06:24 Okay.

06:25 So let's actually do this method.

06:28 This is actually 200 qm plus 2 ,800 qs minus the cm plus cm plus cm plus cs.

06:39 I believe we already calculated the cm plus cs is the c up here.

06:44 So that's a c.

06:46 So i can write it there.

06:51 So this is 2 ,000 qm plus 2 ,800 qs, and then minus 150 qm cube minus qs square.

07:07 Okay, and then minus 3 ,000.

07:13 So this is the profit function.

07:15 And you want to maximize this guy, okay? again, with the constraint that qm plus qs is equal to 400.

07:24 So to illustrate the other method, let's try the substitution method, which i don't really like it, but anyway, that's, it's a good example.

07:37 So let's say that i just want qs.

07:43 Okay, so 2 ,400 minus qs plus 2 ,400 minus qs plus 2 ,000...

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