OM/Claudio_Maggioni_5/Claudio_Maggioni_5.md

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title: Homework 5 -- Optimization Methods author: Claudio Maggioni header-includes:

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Exercise 2

Exercise 2.1

The resulting MATLAB plot of each constraint and of the feasible region is shown below:

Exercise 2.3

We then compute the objective function value for each basic feasible point found, The smallest objective value will correspond with the constrained minimizer problem solution.

x_1 = \begin{bmatrix}0\\0\end{bmatrix} \;\;\; f(x_1) = 4 \cdot 0 + 3 \cdot 0 =
0$$$$
x_2 = \frac12 \cdot \begin{bmatrix}0\\3\end{bmatrix} \;\;\;
f(x_2) = 4 \cdot 0 + 3 \cdot \frac32 = \frac92$$$$
x_3 = \frac{1}{13} \cdot \begin{bmatrix}3\\24\end{bmatrix} \;\;\; f(x_3) = 4
\cdot \frac{3}{13} + 3 \cdot \frac{24}{13} = \frac{84}{13}$$$$
x_4 = \frac12 \cdot \begin{bmatrix}3\\2\end{bmatrix} \;\;\; f(x_4) = 4 \cdot
frac32 + 3 \cdot 1 = 9$$$$
x_5 = \begin{bmatrix}2\\0\end{bmatrix} \;\;\; 4 \cdot 2 + 1 \cdot 0 = 8$$

Therefore, $x^* = x_1$ is the global constrained minimizer with $\lambda^* = \lambda_1 = NaN$ as
the slack variable value.