Contoh Soal 4 Program Linear (Matematika Wajib Kelas XI)

 $\begin{array}{ll}\\ 16.&\textrm{Pada pertidaksamaan}\\ & 2y\geq x\: ;\: y\leq 2x\: ;\: 2y+x\leq 20\: ;\: x+y\geq 9\\ &\textrm{Nilai maksimum untuk}\: \: \color{red}3y-x\: \: \color{black}\textrm{dicapai saat}\: ....\\\\ \end{array}$

$.\: \: \: \quad\begin{array}{ll}\\ \textrm{a}.&\textrm{P}\\ \textrm{b}.&\textrm{Q}\\ \color{red}\textrm{c}.&\textrm{R}\\ \textrm{d}.&\textrm{S}\\ \textrm{e}.&\textrm{T}\\ \end{array}$

$.\: \: \: \quad\begin{aligned}&\textrm{Jawab}:\quad \color{red}\textbf{c}\\ &\color{blue}\textrm{Dengan membuat garis(selidik)}:f(x,y)=3y-x\\ &\color{blue}\textrm{digeser dari bawah ke atas, maka akan didapatkan}\\ &\color{blue}\textrm{titik sudut(verteks) yang diinginkan}\\ & \end{aligned}$

$\begin{array}{ll}\\ 17.&\textrm{Nilai minimum dari}\: \: -2x+4y+6\\ &\textrm{untuk}\: \: x\: \: \textrm{dan}\: \: y\: \: \textrm{yang memenuhi}\\ &\begin{cases} 2x+y-20 &\leq 0 \\ 2x-y+10&\geq 0 \\ x+y-5 &\geq 0 \\ x-2y-5 &\leq 0 \\ x\geq 0 & \\ y\geq 0 & \end{cases}\\ &\textrm{adalah}\: ....\\ &\begin{array}{ll}\\ \textrm{a}.&-14\\ \textrm{b}.&-11\\ \textrm{c}.&-9\\ \textrm{d}.&-6\\ \color{red}\textrm{e}.&-4\\ \end{array}\\ \end{array}$

$.\: \: \: \quad\begin{aligned}&\textrm{Jawab}:\quad \color{red}\textbf{e}\\ &\color{blue}\textrm{Diketahui fungsi objektif}:f(x,y)=-2x+4y+6\\ &\color{blue}\textrm{dan kendala-kendalanya}\\ &\begin{cases} 2x+y-20 &\leq 0 \\ 2x-y+10&\geq 0 \\ x+y-5 &\geq 0 \\ x-2y-5 &\leq 0 \\ x\geq 0 & \\ y\geq 0 & \end{cases}\\ &\color{blue}\textrm{maka daerah penyelesaiannya adalah}: \end{aligned}$

$.\: \: \: \quad\begin{aligned}&\color{blue}\textrm{Dan persamaan garisnya adalah}\\ &\begin{cases} \textrm{L}_{1}\equiv &2x+y=20 \\ \textrm{L}_{2}\equiv &2x-y=-10\\ \textrm{L}_{3}\equiv &x+y=5 \\\textrm{L}_{4} \equiv &x-2y=5 \end{cases}\\ &\color{blue}\textrm{Perpotongan untuk garis}\: \: \textrm{L}_{1}\&\textrm{L}_{2}\\ &\textrm{akan didapatkan titik C}\: \left ( \displaystyle \frac{5}{2},15 \right )\\ &\color{blue}\textrm{Perpotongan untuk garis}\: \: \textrm{L}_{1}\&\textrm{L}_{4}\\ &\textrm{akan didapatkan titik B}\: \left ( \displaystyle 9,2 \right )\\ \end{aligned}$

$.\: \: \: \quad\begin{array}{|l|}\hline \begin{aligned}\color{magenta}\textrm{untuk}\: &\color{magenta}\textrm{mendapatkan titik potong}\\ \textrm{garis}:\: &\color{blue}\textrm{L}_{1}\&\textrm{L}_{2}\: \: \color{black}\textrm{adalah sebagai berikut}:\\ &\begin{cases} 2x+y & =20 \\ 2x-y & =-10 \end{cases}\\ &-------\: \: .^{-}\\ &\: \: \: \quad\quad\quad2y=30\\ &\: \qquad\qquad y=15\Rightarrow x=\displaystyle \frac{5}{2}\\ \textrm{Sehing}&\textrm{ga didapatkan titik}\: \: \left ( \displaystyle \frac{5}{2},15 \right ) \end{aligned} \\\hline \begin{aligned}\color{magenta}\textrm{untuk}\: &\color{magenta}\textrm{mendapatkan titik potong}\\ \textrm{garis}:\: &\color{blue}\textrm{L}_{1}\&\textrm{L}_{4}\: \: \color{black}\textrm{adalah sebagai berikut}:\\ &\begin{cases} 2x+y & =20 \\ x-2y & =5 \end{cases}\\ &\begin{cases} 4x+2y & =40 \\ x-2y & =5 \end{cases}\\ &-------\: \: .^{+}\\ &\qquad\quad\quad5x=45\\ &\: \: \qquad\qquad x=9\Rightarrow y=2\\ \textrm{Sehing}&\textrm{ga didapatkan titik}\: \: \left ( 9,2 \right ) \end{aligned}\\\hline \end{array}$

$.\: \: \: \quad\textrm{Selanjutnya}$

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

$\begin{array}{ll}\\ 18.&\textrm{Nilai minimum}\: \: f(x,y)=3+4x-5y\\ &\textrm{untuk}\: \: x\: \: \textrm{dan}\: \: y\: \: \textrm{yang memenuhi}\\ &\begin{cases} -x+y &\leq 1 \\ x+2y&\geq 5 \\ 2x+y &\leq 10 \\ \end{cases}\\ &\textrm{adalah}\: ....\\ &\begin{array}{ll}\\ \textrm{a}.&-19\\ \textrm{b}.&-6\\ \color{red}\textrm{c}.&-5\\ \textrm{d}.&-3\\ \textrm{e}.&23\\ \end{array} \\ \end{array}$

$.\: \: \: \quad\begin{aligned}&\textrm{Jawab}:\quad \color{red}\textbf{c}\\ &\color{blue}\textrm{Diketahui fungsi objektif}:f(x,y)=3+4x-5y\\ &\color{blue}\textrm{dan kendala-kendalanya}\\ &\begin{cases} -x+y &\leq 1 \\ x+2y&\geq 5 \\ 2x+y &\leq 10 \\ \end{cases}\\ &\color{blue}\textrm{maka daerah penyelesaiannya adalah}: \end{aligned}$

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

$\begin{array}{ll}\\ 19.&\textrm{Fungsi}\: \: f(x,y)=10x+15y\\ &\textrm{untuk}\: \: x\: \: \textrm{dan}\: \: y\: \: \textrm{yang memenuhi}\\ &\begin{cases} x &\geq 0 \\ y&\geq 0 \\ x &\leq 800 \\ x&+\: \: y\: \: \leq 1000\\ \end{cases}\\ &\textrm{mempunyai nilai maksimum}\: ....\\ &\begin{array}{ll}\\ \textrm{a}.&9.000\\ \textrm{b}.&11.000\\ \color{red}\textrm{c}.&13.000\\ \textrm{d}.&15.000\\ \textrm{e}.&16.000\\ \end{array} \\ \end{array}$

$.\: \: \: \quad\begin{aligned}&\textrm{Jawab}:\quad \color{red}\textbf{c}\\ &\color{blue}\textrm{Diketahui fungsi objektif}:f(x,y)=10x+15y\\ &\color{blue}\textrm{dan kendala-kendalanya}\\ &\begin{cases} x &\geq 0 \\ y&\geq 0 \\ x &\leq 800 \\ x&+\: \: y\: \: \leq 1000\\ \end{cases}\\ &\color{blue}\textrm{maka daerah penyelesaiannya adalah}: \end{aligned}$

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

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

$.\: \: \: \quad\begin{aligned}&\textrm{Jawab}:\quad \color{red}\textbf{d}\\ &\color{blue}\textrm{Diketahui fungsi objektif}:f(x,y)=4x+5y\\ &\color{blue}\textrm{dan kendala-kendalanya}\\ &\begin{cases} x&\geq 0 \\ y&\geq 0 \\ (&2x+y-4\: )(\: 2x+3y-6\: )\: \: \leq 0\\ \end{cases}\\ &\color{blue}\textrm{maka daerah penyelesaiannya adalah}: \end{aligned}$

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

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