I like it when you said "What makes this perhaps the most difficult step of all of these optimization questions, is figuring out how and when are they both zero..." Why must it literally take two hours of browsing, just to find a single video that doesn't only spend no more than two seconds on the trickiest part? Geez, Louise!
Mitch, can you please tell me what did you use to record this video? Are you using a surface pro, or some other stylus enabled device, and some screen capture program?
This was done by using Camtasia for Mac to record my screen. The slides are being displayed using the Mac app Skim, and I'm writing using a Wacom Bamboo tablet.
I'm not sure I understand your question. We only need to address y=0 when we're in what I called "Case B", which is where we know x-y-1=0. That forces that x=1 if y=0. You seem to be suggesting that (0,0) should be a critical point, but it clearly doesn't make the partial derivative of f with respect to y 0, so it can't be a critical point.
Okay, I know how to solve a critical point with one variable, but my class is not showing us how to solve with 2 variables. Talk about education system fail. Also, aren't you supposed to just find the derivative of the function, set it to 0, and that is how you find the critical point? What the hell was everything else?
This is a topic for *multivariable* calculus. If you're not taking multivariable calculus (sometimes called Calculus III), you would not be encountering functions of two variables. Based on your second paragraph, I suspect you're in a single-variable (Calculus I) class, since there's no such thing as "the derivative" of a function of two or more variables, only partial derivatives and directional derivatives.
Excellent video! Thanks for explaining everything so clearly!
This is the only video that I've been cleared on more than that of the organic chemistry tutor. Kudos to u
I like it when you said "What makes this perhaps the most difficult step of all of these optimization questions, is figuring out how and when are they both zero..." Why must it literally take two hours of browsing, just to find a single video that doesn't only spend no more than two seconds on the trickiest part? Geez, Louise!
Thank you for the clear and informative video!
Tommorow it's me Vs calculus.Thanks sir
Thank you. You really helped me understand how to find critical points.
Such a helpful video thank u sir
Thanks for the good explanation Mr Keller
Thank you very much
This is absolutely helpful, thanks a lot.
Love this video!
Thank you! Thank you! Thank you so much! Much obliged!
MY LIFE SAVIOUR ! THANKYOU SO MUCH
This was helpful! Thanks!
Thanks alot for this.
Pls aren't the conclusions less than 0 local minima
Thank you.
Thanks 🙏 🙏🙏
Mitch, can you please tell me what did you use to record this video? Are you using a surface pro, or some other stylus enabled device, and some screen capture program?
This was done by using Camtasia for Mac to record my screen. The slides are being displayed using the Mac app Skim, and I'm writing using a Wacom Bamboo tablet.
thank you mitch
you are awesome
Muito Bom
Why did you ignore when y=0 using 6x^2-6xy-6x gives us x= 0 and x=2. You ignored x=0.
Thanks
I'm not sure I understand your question. We only need to address y=0 when we're in what I called "Case B", which is where we know x-y-1=0. That forces that x=1 if y=0. You seem to be suggesting that (0,0) should be a critical point, but it clearly doesn't make the partial derivative of f with respect to y 0, so it can't be a critical point.
@@mitchkeller
Am confused in that point too
Yeah i hope my exam questions will not be like this 😂😂
Okay, I know how to solve a critical point with one variable, but my class is not showing us how to solve with 2 variables. Talk about education system fail.
Also, aren't you supposed to just find the derivative of the function, set it to 0, and that is how you find the critical point? What the hell was everything else?
This is a topic for *multivariable* calculus. If you're not taking multivariable calculus (sometimes called Calculus III), you would not be encountering functions of two variables. Based on your second paragraph, I suspect you're in a single-variable (Calculus I) class, since there's no such thing as "the derivative" of a function of two or more variables, only partial derivatives and directional derivatives.
Well, according to my calculations the derivative with respect to x should be 6x^2 - 6xy - 6x + 3, and with respect to y - without "+3"
To have a constant term of 3 in the partial derivative with respect to x, you'd need a 3x term in the function. There isn't one.
+Mitch Keller Oh, you're right, sorry! I have re-writed on my paper with '3x' instead of '3y', my bad!