Question

Optimization problem... traditional method or Lagrange multipliers? And how do you find the critical points on the boundary? (Photo of the problem in the comments)

Answer 1

on the line \(y = 10 - x^2\), to do it the single variable way, sub \( y = 10 - x^2\) into f(x,y) making it f(x) and just do single var calculus in usual way

for the mutiplier, your constraint is \(g(x,y) = y + x^2\) and so \(\nabla f(x,y) = \lambda g(x,y)\)

is that what you were asking?

Answer 2

typo

\[\nabla f(x,y) = \lambda \color{red}{\nabla}g(x,y)\]

Answer 3

How do you know that the constraint is \[y+x^2\]?

Answer 4

i don't know how much you know of the derivation but the constraint is written as a level surface, that is \(g(x,y) = const.\)

so from \(y = 10 - x^2\), you can say that \(g(x,y) = y + x^2 - 10 = 0\) or even \(g(x,y) = y + x^2 = 10\) .

i dropped the 10 because it falls out in the calculus.

Answer 5

$$f(x,y)=(x-1)(y-2)$$
first look at our critical points where \(\nabla f=\langle y-2,x-1\rangle\) vanishes so we have \((1,2)\) which is within our domain. now let's look at the hessian: $$H=\det\begin{bmatrix}\frac{\partial^2f}{\partial x^2}&\frac{\partial^2f}{\partial x\partial y}\\\frac{\partial^2f}{\partial x\partial y}&\frac{\partial^2f}{\partial y^2}\end{bmatrix}=\det\begin{bmatrix}0&1\\1&0\end{bmatrix}=-1$$ergo we have a saddle point here. by the extreme value theorem we know to look at the boundaries of our domain.

consider the left edge of our domain \(x=0,0\le y\le 10\) so: $$f(0,y)=2-y\\\implies f(0,0)=2, f(0,10)=-8$$
similarly, we take for the bottom edge \(y=0, 0\le x\le\sqrt{10}\) so $$f(x,0)=2-2x\\\implies f(\sqrt{10},0)=2-2\sqrt{10}\approx -4.32$$
and lastly we consider for the top right edge \(0\le x\le\sqrt{10},y=10-x^2\) so $$f(x,10-x^2)=(x-1)(8-x^2)$$where the critical values are clearly $$\frac{d}{dx}(x-1)(8-x^2)=0\\0=(8-x^2)-2x(x-1)\\3x^2-2x-8=0\\\implies x=\frac16(2\pm10)=-\frac43,2$$clearly only \((2,6)\) is in range which gives $$f(2,6)=4$$
so our minimum is \(f=-8\) at \((0,10)\) and our maximum is \(f=4\) at \((2,6)\)

Answer 6

@IrishBoy123 think about what the lagrangian method actually does and I believe you'll find that your solution using the constraint curve \(x^2+y=10\) only finds the maximum but fails to deliver the minimum; it happens to get lucky that the maximum actually lies on that boundary of the domain but it would obviously fail to determine an extreme value on other boundaries or even in the interior of the domain

Answer 7

in order to use Lagrangian multipliers here we need to actually extend to the more general Kuhn-Tucker conditions; we want \(x^2+y\le 10, x\ge 0, y\ge 0\) so we introduce the multipliers \(\mu_i\) for our inequality constraints to form a Lagrangian objective like so: $$\Lambda(x,y)=(x-1)(y-2)+\mu_1(10-x^2-y)+\mu_2x+\mu_3y$$it follows that our maximal value will occur where \(\Lambda\) has a critical point and the multipliers satisfy the following constraints $$\mu_1(10-x^2-y)=0\quad\mu_2 x=0\quad\mu_3y=0$$i.e. either the constraint has some slack and we don't need a violation penalty (\(\mu_i=0\)) or the constraint is tight and we do.

our critical points clearly satisfy:
$$\frac{\partial\Lambda}{\partial x}=0\\\implies y-2-2x\mu_1+\mu_2=0\\\frac{\partial\Lambda}{\partial y}=0\\\implies x-1-\mu_1+\mu_3=0$$
so we get the following system: $$-2\mu_1 x+y=2-\mu_2\\x=1+\mu_1+\mu_3\\\mu_1(10-x^2-y)=0\\\mu_2x=0\\\mu_3y=0$$

Answer 8

oops, that should read \(x=1+\mu_1-\mu_3\)

Answer 9

the solutions we get are listed here (\(a=\mu_1, b=\mu_2, c=\mu_3\)):

http://www.wolframalpha.com/input/?i=y-2-2ax%2Bb%3D0,+x-1-a%2Bc%3D0,+a(10-x%5E2-y)%3D0,+bx%3D0,+cy%3D0

which gives our minimum and maximum readily

Answer 10

brilliant stuff oldrin, thank you!