math119: add optimisation constraints

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eggy 2023-01-28 19:21:43 -05:00
parent da0bae4c69
commit 828a653740

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@ -288,3 +288,37 @@ Local maxima tend to be **concave down** while local minima are **concave up**.
a. If $f_{xx}(P_0)<0$, the point is a maximum otherwise it is a minimum
3. If it is less than zero, it is a saddle point — otherwise the test is inconclusive and you must use your eyeballs
### Optimisation with constraints
If there is a limitation in optimising for $f(x,y)$ in the form $g(x,y)=K$, new critical points can be found by setting them equal to each other, where $\lambda$ is the **Lagrange multiplier** that determines the rate of increase of $f$ with respect to $g$:
$$\nabla f = \lambda\nabla g, g(x,y)=K$$
If possible, $\nabla g=\vec 0, g(x,y)=K$ should also be tested.
!!! example
If $A(x,y)=xy$, $g(x,y)=K: x+2y=400$, and $A(x,y)$ should be maximised:
\begin{align*}
\nabla f &= (y, x) \\
\nabla g &= (1, 2) \\
(y, x) &= \lambda (1, 2) \\
\begin{cases}
y &= \lambda \\
x &= 2\lambda \\
x + 2y &= 400 \\
\end{cases}
\\
\\
\therefore y=100,x=200,A=20\ 000
\end{align*}
This applies equally to higher dimensions and constraints by adding a new term for each constraint. Given $f(x,y,z)$ with constraints $g(x,y,z)=K$ and $h(x,y,z)=M$:
$$\nabla f=\lambda_1\nabla g + \lambda_2\nabla h$$
### Absolute extrema
- If end points exist, those should be added
- If no endpoints exist and the limits go to $\pm\infty$, there are no absolute extrema