A visual solution

# Draw the objective function contour line where z=ax+by
z0=21
n<-c(3,2)
x<-(0:90)/10
yz<-(z0-n[1]*x)/n[2]
plot(x,yz,type='l',lwd=3,col="red",xlim=c(0,10),ylim=c(0,10))
arrows(x[20], yz[20], x[20]+n[1], yz[20]+n[2],col="red",lwd=3)
text(x[20]+n[1], yz[20]+n[2]," n",adj=0,cex=3)

z0=27
n<-c(3,2)
x<-(0:90)/10
yz<-(z0-n[1]*x)/n[2]
lines(x,yz,type='l',lwd=3,col="red")

                                        # Draw a half-plane a'x<=b
a1<-c(1,1)
b1<-10
y1<-(b1-a1[1]*x)/a1[2]
lines(x,y1,type='l',lwd=3,col="blue")
x1<-(5:60)/10
y1<-(b1-a1[1]*x1)/a1[2]
xl1<-c(rbind(x1,x1,rep(NA,length(x1))))
yl1<-c(rbind(y1,y1-1,rep(NA,length(x1))))
lines(xl1,yl1)
text(5,5,"x+y=10",adj=0,cex=2)

                                        # Draw a half-plane a'x<=b
a2<-c(1,-1)
b2<- 5
y1<-(b2-a2[1]*x)/a2[2]
lines(x,y1,lwd=3,col="blue")
x1<-(50:70)/10
y1<-(b2-a2[1]*x1)/a2[2]
xl2<-c(rbind(x1,x1,rep(NA,length(x1))))
yl2<-c(rbind(y1,y1+1,rep(NA,length(x1))))
lines(xl2,yl2)
text(8.3,3.0,"x-y=5",adj=0,cex=2)

lines(c(0,10),c(0,0))
xgrd1<-(0:45)/10
xgrd<-rep(xgrd1,rep(3,length(xgrd1)))
ygrdl<-rep(0,length(xgrd1))
ygrdu<-rep(1,length(xgrd1))
ygrdNA<-rep(NA,length(xgrd1))
ygrd<-c(rbind(ygrdl,ygrdu,ygrdNA))
lines(xgrd,ygrd)


lines(c(0,0),c(0,10))
ygrd1<-(0:90)/10
ygrd<-rep(ygrd1,rep(3,length(ygrd1)))

xgrdl<-rep(0,length(ygrd1))
xgrdu<-rep(1,length(ygrd1))
xgrdNA<-rep(NA,length(ygrd1))
xgrd<-c(rbind(xgrdl,xgrdu,xgrdNA))
lines(xgrd,ygrd)
$$
min_x c'x
$$
w.r.t.
$$
Ax \leq b
$$
Explanation
GeoGebra is an excellent tool to draw the figures \verb#https://www.youtube.com/wat[…]940t-PA&feature=relmfu#

{\bf Example (from a 2004 exam in Iceland)}
Draw a figure and use it to find the values, $x$ and $y$, which minimize
$z= 2x-y$ with respect to
\begin{eqnarray*}
4x+ 3y & \leq & 12 \\
-x + y & \leq & 2 \\
 x + 2y & \geq & 2 \\
 x & \geq & 0 \\
      y & \geq & 0
\end{eqnarray*}
Details
If there are only two variable one can draw all the constraints, find the set of admissible solutions and inspect visually how the function changes within the solution set.
Examples
{\bf Dæmi:}
Látum $x$ tákna magn (kg) af fóðurgerð A og $y$ magn af gerð B sem
á að nota í dagsskammt.

Úr því fóðurgerð A kostar 10 kr/kg og B kostar 20 kr/kg er verð
blöndunnar $z=10x+20y$.

Orkuinnihald dagsskammtsins byggist á því að
orkuinnihald A er 500 kaloríur (Kcal) á kg og í B eru 4000 Kcal/kg
þannig að dagsskammturinn er með $500x+4000y$.

Í A eru 200g prótíns á hvert kg, en
í B eru 100g/kg, þannig að alls eru $200x+100y$ í skammtinum.

Til að lágmarka kostnað en ná a.m.k. 2000 Kcal/dag og 400g af prótíni á dag
viljum við finna minnsta mögulega gildi á $z=10x+20y$ þó þannig að
$500x+4000y\geq 2000$ og $200x+100y \geq 400$.

Oftast er þetta orðað þannig:
\begin{eqnarray*}
min \ z&=&10x+20y \\
\textrm{m.t.t. } \\
500x+4000y&\geq &2000 \\
200x+100y &\geq &400
\end{eqnarray*}

GeoGebra is an excellent tool to draw the figures \verb#https://www.youtube.com/wat[…]940t-PA&feature=relmfu#

{\bf Example (from a 2004 exam in Iceland)}
Draw a figure and use it to find the values, $x$ and $y$, which minimize
$z= 2x-y$ with respect to
\begin{eqnarray*}
4x+ 3y & \leq & 12 \\
-x + y & \leq & 2 \\
 x + 2y & \geq & 2 \\
 x & \geq & 0 \\
      y & \geq & 0
\end{eqnarray*}

Justify the answer.