@17:41 "We're gonna be off by a negative here." The minus sign is superfluous. The constraint, g(x,y), is a homogeneous equation [it's equal to zero]. As such it literally does not matter whether you add or subtract it from the objective function when forming the Lagrangian function, F(x,y,λ). Most mathematicians would write F(x,y,λ) = f(x,y) - λ·g(x,y), where g(x,y) = 100x + 200y - 30000 = 0. Economists have a weird obsession with non-standard notation. They think it enhances job security or something. To an economist the Lagrangian is typically written as F(x,y,λ) = f(x,y) + λ·g(x,y), where g(x,y) = 30000 - 100x - 200y = 0. In this case the minus sign that would typically be in front of λ is distributed across the terms of g(x,y). But even this doesn't make any difference in the solution to the problem because g(x,y) = 0 and because nobody cares about the sign of λ; it's only a parameter guaranteeing that grad f is parallel to grad g. You didn't make a mistake in setting the problem up the way you did in the first place. 😎
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Thanks! But you only solved for critical values, and not optimal quantity.
@17:41 "We're gonna be off by a negative here."
The minus sign is superfluous. The constraint, g(x,y), is a homogeneous equation [it's equal to zero]. As such it literally does not matter whether you add or subtract it from the objective function when forming the Lagrangian function, F(x,y,λ).
Most mathematicians would write F(x,y,λ) = f(x,y) - λ·g(x,y), where g(x,y) = 100x + 200y - 30000 = 0. Economists have a weird obsession with non-standard notation. They think it enhances job security or something. To an economist the Lagrangian is typically written as F(x,y,λ) = f(x,y) + λ·g(x,y), where g(x,y) = 30000 - 100x - 200y = 0. In this case the minus sign that would typically be in front of λ is distributed across the terms of g(x,y). But even this doesn't make any difference in the solution to the problem because g(x,y) = 0 and because nobody cares about the sign of λ; it's only a parameter guaranteeing that grad f is parallel to grad g. You didn't make a mistake in setting the problem up the way you did in the first place. 😎
Thank you very much, your work is appreciated :)
Dont you have to take the negative of the objective function since this is a maximization problem?
from scipy.optimize import minimize as mimi
sol=mimi(lambda a:2*a[0]*a[1],x0=[1,1],constraints=[{'type':'eq','fun':lambda a:a[0]+a[1]-6}])
print(sol)
message: Optimization terminated successfully
success: True
status: 0
fun: 18.000000357627883
* x: [ 3.000e+00 3.000e+00]*
nit: 3
jac: [ 6.000e+00 6.000e+00]
nfev: 10
njev: 3
Explained quite well. Where can one get the book?
such a good and funny professor..Thanks a lot
This is really helpful , thank you
What textbook are you using to get this theorem?
Please do a video on second order condition?
Thank you very much. Not making it formal made it interesting. Imagine! Maths? Interesting?
Q= ALαKβ
C = wL + rK
F(L,K, λ) = ALαKβ + λ(C-wL-rK)
Where r=5; w=10; α=1/2; β=1/2;
How can I find L,K,Q, and λ?
You never explained why (3,3) was a point of maximum and not of minimum. You must explain the sufficient conditions as well.
Thank you very much
what is is the maximum value of the second problem?
Thanks a lot.
A very Great piece.
😄😄😄😄😄thank you ma'am, yo explain in so well!!
Thank you love you hug you and appreciate you
thank you
don't we need to do the second derivative test with the bordered hessian matrix at the end of each exercise ?
Are u mad women frist you learn
Absolutely loved the video
Tysm....this video proves to b very useful for me...😊
Very well explained
so helpful thank you
What software do you use to write?
Did u know it?
Tysm❤️