thank you thank you thank you thank you thank you a millions times thank you. so grateful to have found this channel. studying micro from a textbook is literally impossible
@@KatherineSilzCarson so for the short run i only solve for 1 of the xs? And for example if i chose to solve for x1 in the short run , is it the same way as we solved for x1 in the long run?
@@LR2yt Yes. In the short run, at least one input is fixed, so you would only solve for the input that is variable, and treat the fixed inputs as constants when you take the derivative. You will use the same methodology for the short run and the long run, but the math will be slightly different for the long run, because you are differentiating with respect to multiple inputs (at least 2), and you have a system of n equations and n unknowns (where n is the number of inputs), whereas in the short run, you are (usually) only differentiating with respect to a single input and solving a single equation with a single unknown. Hope this helps!
Hi Sagar - I have only made videos for those chapters that I include on my syllabus. I can certainly add your request to my list for the future. Thank you for watching!
9:06 "I will use the first equation to solve for x2 to the 2/3", but the first equation doesn't have x2^2/3. It only has x1^-2/3 and x2^1/3 can you please help me?
@Simone, you are correct, I solved for x2 to the 1/3, but then if I square that, I get the solution for x2 to the 2/3. That is useful because x2 to the 2/3 is in the second equation, so once I have an expression for that in terms of x1, I can just substitute directly and the 2/3 power magically goes away. Hope this helps!
Mis Carson sorry but if you work with wi / w2 you will enter x2 / x1 soon the optimal value will be x1 = p ^ 2/3. (w2 / w1) ^ 1/2 which remains if w2 increases x1 increases and if w1 increases x1 decreases
Hey Ms.Carson, I have a quick question about the methodology. I am using this same method that you have used in the video yet I am still not sure what method it is. At the beginning i thought it was the lagrange multiplier. Could you let me know what the method is called please
Mohamed - it is just straight maximization. Since there is no constraint in this problem, there is no need to use the Lagrange multiplier method. You use that in constrained maximization problems (like utility maximization problems). Hope this helps! Thank you for watching!
Another question about an equation! :) How does it work at 6:15 when taking X2^1/3 out... Why does the W1 get caught in the division and turn into 3W. I have looked at it over and over again... but I do not get it. Thank you in advance! :) Also... maybe it might be good for such equations to be at the middle of the page/ top of the page... Lay-out wise. Because now, when I pause the it to take extra time to figure it out, the red 'play' line goes right through it... Makes it harder to read it.
Hi Simone - here is how it works. The goal of doing this is to get x2 by itself and everything else on the right-hand side of the equation. Step 1 is to add w1 to both sides, so there will be no w's left on the left hand side and w1 will be on the right hand side. Then, I want to get the 3 out of the denominator, so I multiply both sides of the equation by 3. This gets rid of the 3 on the left-hand side, and the right-hand side becomes 3w1. Hope this helps! Also, thank you for your comment about the red line cutting through the equation when you pause the video. No one has pointed this out to me before, and I see how this can impede your understanding. I'll put this on my list of things to fix. Thank you for watching!
Yes Madhav - you are correct. The x1 should be a w2. That is a typo that I hadn't noticed before. Thank you for catching it! And thank you for watching!
What I did was use the first equation to solve for x2^1/3. Then, I square that to get an expression for x2^2/3. The reason why I do that is the second equation contains x2^2/3. Thus, if I can get an expression for x2^2/3, I can substitute that expression directly into the second equation, and the ugly 2/3 exponent magically disappears. Hope this helps!
Miss carson, what would be the best way to contact you if I still have further questions? I don't see an email here? Would greatly appreciate your immediate response thank you.
Hi! I would prefer that you post your questions in the comments, so that everyone who watches can see the answer. If you have questions, chances are that other people do too, so it is helpful to have the answers to those questions visible to all. Thank you for watching!
thank you thank you thank you thank you thank you a millions times thank you. so grateful to have found this channel. studying micro from a textbook is literally impossible
Glad you have found it to be helpful. Thank you for watching!
Your video is easier to understand than what my prof teaches in online class. Keep up the good work! Love from Philippines!
I am glad that you are finding these helpful! Thank you for watching!
Great video. In 6:59 it will be clearer if it said the "slope of revenue curve"
Yes - thank you for catching that error!
Thanx a lot... I regret paying for my tution... Love from India...
Thank you for watching!
Hello
Is it possible to make a video on how to calculate the profit max in the short run with cobb douglas ?
The example function in the video in the long-run profit maximization section is a Cobb-Douglas production function. Hope this helps!
@@KatherineSilzCarson so for the short run i only solve for 1 of the xs? And for example if i chose to solve for x1 in the short run , is it the same way as we solved for x1 in the long run?
@@LR2yt Yes. In the short run, at least one input is fixed, so you would only solve for the input that is variable, and treat the fixed inputs as constants when you take the derivative. You will use the same methodology for the short run and the long run, but the math will be slightly different for the long run, because you are differentiating with respect to multiple inputs (at least 2), and you have a system of n equations and n unknowns (where n is the number of inputs), whereas in the short run, you are (usually) only differentiating with respect to a single input and solving a single equation with a single unknown. Hope this helps!
Ma'am why u r not posting videos of other chapters...like uncertainty and all from HL Varian book?
Hi Sagar - I have only made videos for those chapters that I include on my syllabus. I can certainly add your request to my list for the future. Thank you for watching!
9:06 "I will use the first equation to solve for x2 to the 2/3", but the first equation doesn't have x2^2/3. It only has x1^-2/3 and x2^1/3
can you please help me?
@Simone, you are correct, I solved for x2 to the 1/3, but then if I square that, I get the solution for x2 to the 2/3. That is useful because x2 to the 2/3 is in the second equation, so once I have an expression for that in terms of x1, I can just substitute directly and the 2/3 power magically goes away. Hope this helps!
@@KatherineSilzCarson Thank you so much Katherine! After all these years still answering questions, you're the best!
Mis Carson sorry but if you work with wi / w2 you will enter x2 / x1 soon the optimal value will be
x1 = p ^ 2/3. (w2 / w1) ^ 1/2 which remains if w2 increases x1 increases and if w1 increases x1 decreases
Jose - yes - I thought that's what I said. If there's an error (I'll check), thank you for catching!
@@KatherineSilzCarson I'm an electrical engineer and I' ve starting to lear from your amazing class
@@josecarlosgrizendi553 I'm glad you're finding it useful. Thank you for watching!
Hey Ms.Carson, I have a quick question about the methodology. I am using this same method that you have used in the video yet I am still not sure what method it is. At the beginning i thought it was the lagrange multiplier. Could you let me know what the method is called please
Mohamed - it is just straight maximization. Since there is no constraint in this problem, there is no need to use the Lagrange multiplier method. You use that in constrained maximization problems (like utility maximization problems). Hope this helps! Thank you for watching!
@@KatherineSilzCarson Oh yes thank you! it is already assumed that the constraint is within the production function right? F(x1.x2)
@@mhdlamin - yes, the production function is basically the constraint.
Another question about an equation! :) How does it work at 6:15 when taking X2^1/3 out... Why does the W1 get caught in the division and turn into 3W. I have looked at it over and over again... but I do not get it. Thank you in advance! :)
Also... maybe it might be good for such equations to be at the middle of the page/ top of the page... Lay-out wise. Because now, when I pause the it to take extra time to figure it out, the red 'play' line goes right through it... Makes it harder to read it.
Hi Simone - here is how it works. The goal of doing this is to get x2 by itself and everything else on the right-hand side of the equation. Step 1 is to add w1 to both sides, so there will be no w's left on the left hand side and w1 will be on the right hand side. Then, I want to get the 3 out of the denominator, so I multiply both sides of the equation by 3. This gets rid of the 3 on the left-hand side, and the right-hand side becomes 3w1. Hope this helps!
Also, thank you for your comment about the red line cutting through the equation when you pause the video. No one has pointed this out to me before, and I see how this can impede your understanding. I'll put this on my list of things to fix. Thank you for watching!
@@KatherineSilzCarson Thank you for the quick reply! I do see it/ get it now. Thanks a lot!
Love from India. Hope u put up more videos.
Thank you for watching Abdul!
At 11:27 , Shouldn't the equation for factor demand for x1 have w1^2w2 in the denominator?
Yes Madhav - you are correct. The x1 should be a w2. That is a typo that I hadn't noticed before. Thank you for catching it! And thank you for watching!
Hi, at 9.16 where you wrote X2^2/3. and then you squared it all, I don't understand the power of 2/3. can you please explain that
What I did was use the first equation to solve for x2^1/3. Then, I square that to get an expression for x2^2/3. The reason why I do that is the second equation contains x2^2/3. Thus, if I can get an expression for x2^2/3, I can substitute that expression directly into the second equation, and the ugly 2/3 exponent magically disappears. Hope this helps!
Miss carson, what would be the best way to contact you if I still have further questions? I don't see an email here? Would greatly appreciate your immediate response thank you.
Hi! I would prefer that you post your questions in the comments, so that everyone who watches can see the answer. If you have questions, chances are that other people do too, so it is helpful to have the answers to those questions visible to all. Thank you for watching!