Órla Mannion, the method will work for quasilinear convex, cobb-douglass and perfect complements (optimum at x equals to y) because optimum is at the tangency where IDC is tangent to the budget constraint. This method won't work for perfect substitutes, max functions, and quasilinear concave because all 3 of them end up touching axes. Therefore our optimum would be on the boundary. Hope that helps
Hello, John. That sounded very knowledgeable but trust me when I tell you this, I have no idea what quasilinear and max functions and perfect substitutes are😂
Video is very helpful thank you - Just wondering if this applies only to Cobb Douglas utility functions or for perfect compliments and perfect substitutes too? Thanks!
U (X,Y) can be written without the (X,Y), simply as U = *insert an equation* Either form mean the same thing; the parentheses just serve to define the variables that will be used in the equation. So, U(X,Z), can also just be written as U = *insert an equation*, but instead of Y, Z takes its place. Picking variables is arbitrary.
What if the partial derivative of U with respect to Y is equal to 1. So we can’t solve for Y cause it’s cancelled out. As is the case for U = x^0.5 +y?
Órla Mannion, the method will work for quasilinear convex, cobb-douglass and perfect complements (optimum at x equals to y) because optimum is at the tangency where IDC is tangent to the budget constraint. This method won't work for perfect substitutes, max functions, and quasilinear concave because all 3 of them end up touching axes. Therefore our optimum would be on the boundary. Hope that helps
Hello, John. That sounded very knowledgeable but trust me when I tell you this, I have no idea what quasilinear and max functions and perfect substitutes are😂
😘
'Demand for good x is the consumers income divided by this mess in the bottom' hahahahah describes this process perfectly
*"this mess in the denominator"* lmao
Video is very helpful thank you - Just wondering if this applies only to Cobb Douglas utility functions or for perfect compliments and perfect substitutes too? Thanks!
really awesome video, thanks so much for helping me, and many others like me, out
I am glad it helped!
thanks, I really couldn't remember exactly how to do it so this helped a lot :)
this is class cheers. this is 2nd year university stuff lol
thank you this helped a TONNN!!!
Keep creating videos like this. Very helpful.
dang really helped me out in understanding thank you
Thank you it really helped to better my understanding
Thank you for your existence!!
Thanks for the help
Is there a situation where we cannot use martilian demand
Can we figure out if these two goods are related in consumption from from the demand function?
Awesome. Thank you!
You are welcome!
my prof said that for u = x^0.5+y^0.5 the demand curve would be x = i/px (py/px+py)? i don't get it
Extremely helpful. Thanks!
Now what if the utility function is U= xy
same concept same steps
Many thanks. It's helpful.
thanks, very helpful
you're a lifesaver
Why do the .5s cancel at 3:19?
really useful!! However I have a question: why is the utility function u=x^0.5+y^0.5? I hope you can answer!
+Enrique Sanz Posse Because an alternate form of square root is to the power of .5
Thank you very much!!!!
this is sooooooo helpfullll
That is great to hear! Thanks for watching!
what if it is + x just? the derivative will just be one. Can I still go on with this method?
Please can you show me the demand function for good X and Y when the utility function is given as U(X,Y)
U (X,Y) can be written without the (X,Y), simply as U = *insert an equation*
Either form mean the same thing; the parentheses just serve to define the variables that will be used in the equation. So, U(X,Z), can also just be written as U = *insert an equation*, but instead of Y, Z takes its place. Picking variables is arbitrary.
When we first solved for y, was that the demand function for y
Pls ans. I would like to find out about this too.
No you repeat the process for Y that he did for X in order to find the demand function for Y
Very helpful. Thank you
very helpful, thank you so much!
Thanks this really helped
What if the partial derivative of U with respect to Y is equal to 1. So we can’t solve for Y cause it’s cancelled out. As is the case for U = x^0.5 +y?
See here: ruclips.net/video/wKN3qw1uQ54/видео.html
hi, first I thank you for your great and clear explanations . what the difference between this example and the Cobb-Douglas Function ?
Thank you so much for saving me!!
哈哈哈 加油
shouldnt your first y be raised to -2?
very helpful thank you!
thank you! this was very helpful! :)
very helpful
isnt du/dx= 0.5x`-0.5(y`0.5)?
Abim süpersin ya devam et
Thank you!!!!!!
Dear 1sportingclays,
How would this work if the the utility function was u(x1 + x2) = (√x1) + x2 ?
Thank you
I hope you found your answer, 4 years ago :)
GOOD SHIT DAWG
wow it's actually so easy
helpful
At this point economists are just making stuff up 😂
You tell me fam😂. They also debate for hours on end about abstract concepts and take home fat paychecks
😂😂😂😂
cheers mate.
Thank you very much, final who need refreshing after placement!
❤❤❤
Ama yeni formatlarını beğenmedim boyle iyi anlatıyodun daha çok soru çözümü yaparmisin Türkiye'den selamlar
Please make a dumb person version of this as im dumb but want to learn..
I found it hard to follow
tema
Anyone else notice this is wrong ?
why?
why is .5 square root = 1? at 5:13
@@nyanruuu because 0.5 is the same thing as 1/2. And 1/2 square root is 1
@@nyanruuu he just squared everything
@@Anna_Subliminal so it's ½ *squared* = 1 not ½ *_square root_* those are entirely different things