Contoh Soal 4 Program Linear (Matematika Wajib Kelas XI)

 $\begin{array}{l}\\ 16.& \text{Pada pertidaksamaan}\\ & 2y\ge x;y\le 2x;2y+x\le 20;x+y\ge 9\\ & \text{Nilai maksimum untuk}3y-x\text{dicapai saat}....\\ \end{array}$

$.\begin{array}{l}\\ \text{a}.& \text{P}\\ \text{b}.& \text{Q}\\ \text{c}.& \text{R}\\ \text{d}.& \text{S}\\ \text{e}.& \text{T}\end{array}$

$.\begin{array}{rl}& \text{Jawab}:\mathbf{\text{c}}\\ & \text{Dengan membuat garis(selidik)}:f(x,y)=3y-x\\ & \text{digeser dari bawah ke atas, maka akan didapatkan}\\ & \text{titik sudut(verteks) yang diinginkan}\\ & \end{array}$

$\begin{array}{l}\\ 17.& \text{Nilai minimum dari}-2x+4y+6\\ & \text{untuk}x\text{dan}y\text{yang memenuhi}\\ & \{\begin{array}{ll}2x+y-20& \le 0\\ 2x-y+10& \ge 0\\ x+y-5& \ge 0\\ x-2y-5& \le 0\\ x\ge 0& \\ y\ge 0& \end{array}\\ & \text{adalah}....\\ & \begin{array}{l}\\ \text{a}.& -14\\ \text{b}.& -11\\ \text{c}.& -9\\ \text{d}.& -6\\ \text{e}.& -4\end{array}\end{array}$

$.\begin{array}{rl}& \text{Jawab}:\mathbf{\text{e}}\\ & \text{Diketahui fungsi objektif}:f(x,y)=-2x+4y+6\\ & \text{dan kendala-kendalanya}\\ & \{\begin{array}{ll}2x+y-20& \le 0\\ 2x-y+10& \ge 0\\ x+y-5& \ge 0\\ x-2y-5& \le 0\\ x\ge 0& \\ y\ge 0& \end{array}\\ & \text{maka daerah penyelesaiannya adalah}:\end{array}$

$.\begin{array}{rl}& \text{Dan persamaan garisnya adalah}\\ & \{\begin{array}{ll}{\text{L}}_1\equiv & 2x+y=20\\ {\text{L}}_2\equiv & 2x-y=-10\\ {\text{L}}_3\equiv & x+y=5\\ {\text{L}}_4\equiv & x-2y=5\end{array}\\ & \text{Perpotongan untuk garis}{\text{L}}_1\mathrm{\&}{\text{L}}_2\\ & \text{akan didapatkan titik C}\left({\displaystyle \frac{5}{2},15}\right)\\ & \text{Perpotongan untuk garis}{\text{L}}_1\mathrm{\&}{\text{L}}_4\\ & \text{akan didapatkan titik B}\left({\displaystyle 9,2}\right)\end{array}$

$.\begin{array}{|l|}\hline \begin{array}{rl}\text{untuk}& \text{mendapatkan titik potong}\\ \text{garis}:& {\text{L}}_1\mathrm{\&}{\text{L}}_2\text{adalah sebagai berikut}:\\ & \{\begin{array}{ll}2x+y& =20\\ 2x-y& =-10\end{array}\\ & -------{.}^{-}\\ & 2y=30\\ & y=15\Rightarrow x={\displaystyle \frac{5}{2}}\\ \text{Sehing}& \text{ga didapatkan titik}\left({\displaystyle \frac{5}{2},15}\right)\end{array}\\ \begin{array}{rl}\text{untuk}& \text{mendapatkan titik potong}\\ \text{garis}:& {\text{L}}_1\mathrm{\&}{\text{L}}_4\text{adalah sebagai berikut}:\\ & \{\begin{array}{ll}2x+y& =20\\ x-2y& =5\end{array}\\ & \{\begin{array}{ll}4x+2y& =40\\ x-2y& =5\end{array}\\ & -------{.}^{+}\\ & 5x=45\\ & x=9\Rightarrow y=2\\ \text{Sehing}& \text{ga didapatkan titik}(9,2)\end{array}\\ \hline\end{array}$

$.\text{Selanjutnya}$

$.\begin{array}{|ccc|}\hline \text{Verteks}& \text{Nilai}:& \text{Keterangan}\\ & -2x+4y+6& \\ \text{A}(5,0)& -2(5)+4.0+6=-4& \text{Minimum}\\ \text{B}(9,2)& -2(9)+4.2+6=-4& \text{Minimum}\\ \text{C}\left({\displaystyle \frac{5}{2},15}\right)& -2\left({\displaystyle \frac{5}{2}}\right)+4.15+6=61& \text{Maksimum}\\ \text{D}(0,10)& -2.0+4.10+6=46& \\ \text{E}(0,5)& -2.0+4.5+6=26& \\ \hline\end{array}$

$\begin{array}{l}\\ 18.& \text{Nilai minimum}f(x,y)=3+4x-5y\\ & \text{untuk}x\text{dan}y\text{yang memenuhi}\\ & \{\begin{array}{ll}-x+y& \le 1\\ x+2y& \ge 5\\ 2x+y& \le 10\end{array}\\ & \text{adalah}....\\ & \begin{array}{l}\\ \text{a}.& -19\\ \text{b}.& -6\\ \text{c}.& -5\\ \text{d}.& -3\\ \text{e}.& 23\end{array}\end{array}$

$.\begin{array}{rl}& \text{Jawab}:\mathbf{\text{c}}\\ & \text{Diketahui fungsi objektif}:f(x,y)=3+4x-5y\\ & \text{dan kendala-kendalanya}\\ & \{\begin{array}{ll}-x+y& \le 1\\ x+2y& \ge 5\\ 2x+y& \le 10\end{array}\\ & \text{maka daerah penyelesaiannya adalah}:\end{array}$

$.\begin{array}{|ccc|}\hline \text{Verteks}& \text{Nilai}:& \text{Keterangan}\\ & 3+4x-5y& \\ \text{A}(1,2)& 3+4.1-5.2=-3& \\ \text{B}(3,4)& 3+4.3-5.4=-5& \text{Minimum}\\ \text{C}(5,0)& 3+4.5-5.0=23& \text{Maksimum}\\ \hline\end{array}$

$\begin{array}{l}\\ 19.& \text{Fungsi}f(x,y)=10x+15y\\ & \text{untuk}x\text{dan}y\text{yang memenuhi}\\ & \{\begin{array}{ll}x& \ge 0\\ y& \ge 0\\ x& \le 800\\ x& +y\le 1000\end{array}\\ & \text{mempunyai nilai maksimum}....\\ & \begin{array}{l}\\ \text{a}.& 9.000\\ \text{b}.& 11.000\\ \text{c}.& 13.000\\ \text{d}.& 15.000\\ \text{e}.& 16.000\end{array}\end{array}$

$.\begin{array}{rl}& \text{Jawab}:\mathbf{\text{c}}\\ & \text{Diketahui fungsi objektif}:f(x,y)=10x+15y\\ & \text{dan kendala-kendalanya}\\ & \{\begin{array}{ll}x& \ge 0\\ y& \ge 0\\ x& \le 800\\ x& +y\le 1000\end{array}\\ & \text{maka daerah penyelesaiannya adalah}:\end{array}$

$.\begin{array}{|ccc|}\hline \text{Verteks}& \text{Nilai}:& \text{Keterangan}\\ & f(x,y)=10x+15y& \\ \text{A}(800,0)& 800.10+0=8000& \text{Minimum}\\ \text{B}(800,200)& 800.10+15.200=11.000& \\ \text{C}(400,600)& 10.400+15.600=13.000& \text{Maksimum}\\ \text{D}(0,600)& 3+0+15.600=9.000& \\ \hline\end{array}$

$\begin{array}{l}\\ 20.& \text{Nilai maksimum fungsi sasaran}\\ & f(x,y)=4x+5y\\ & \text{untuk}x\text{dan}y\text{yang memenuhi}\\ & \{\begin{array}{ll}x& \ge 0\\ y& \ge 0\\ (& 2x+y-4)(2x+3y-6)\le 0\end{array}\\ & \text{adalah}....\\ & \begin{array}{l}\\ \text{a}.& 11\\ \text{b}.& 12\\ \text{c}.& 16\\ \text{d}.& 20\\ \text{e}.& 24\end{array}\end{array}$

$.\begin{array}{rl}& \text{Jawab}:\mathbf{\text{d}}\\ & \text{Diketahui fungsi objektif}:f(x,y)=4x+5y\\ & \text{dan kendala-kendalanya}\\ & \{\begin{array}{ll}x& \ge 0\\ y& \ge 0\\ (& 2x+y-4)(2x+3y-6)\le 0\end{array}\\ & \text{maka daerah penyelesaiannya adalah}:\end{array}$

$.\begin{array}{rl}\text{untuk}& \text{mendapatkan titik potongnya}\\ & \{\begin{array}{ll}2x+y& =4\\ 2x+3y& =6\end{array}\\ & -------{.}^{-}\\ & -2y=-2\\ & y=1\Rightarrow x={\displaystyle \frac{3}{2}}\\ \text{sehing}& \text{ga akan didapatkan}\\ \text{titik p}& \text{otongnya adalah}:\left({\displaystyle \frac{3}{2},1}\right)\\ \text{Selanj}& \text{utnya, kita dapat menentukan}\\ \text{nilai}& \text{maksimunya dengan bantuan tabel berikut}\end{array}$

$.\begin{array}{|ccc|}\hline \text{Verteks}& \text{Nilai}:& \text{Keterangan}\\ & f(x,y)=4x+5y& \\ (2,0)& 4.2+0=8& \text{Minimum}\\ (3,0)& 4.3+0=12& \\ \left({\displaystyle \frac{3}{2},1}\right)& 4.\left({\displaystyle \frac{3}{2}}\right)+5.1=11& \\ (0,2)& 0+52=10& \\ (0,4)& 0+5.4=20& \text{Maksimum}\\ \hline\end{array}$