Solve the following linear programming problems graphically:
Maximize $Z=-3x+4y$
subject to the constraints $x+2y\le 8,3x+2y\le 12,x\ge 0,y\ge 0$

Hint: First of all draw the inequations given as constraints on the graph and then from the region covered by the constraints take the corner points and then substitute these corner points in Z and see which point is giving you the maximum value. Then the point which is giving you the maximum value is the point at which Z is maximizing.

Complete step-by-step solution -
The constraints that we are subjected to:
$x+2y\le 8,3x+2y\le 12,x\ge 0,y\ge 0$
Using these constraints, we have to maximize $Z=-3x+4y$.
In the below figure, we have drawn the constraints on the graph:

In the above figure, the region covered by EBCD is the intersection of all the constraints that have given in the question $x+2y\le 8,3x+2y\le 12,x\ge 0,y\ge 0$. The point B corresponds to (2, 3).
Now, we are going to substitute the value of points E, B, and C in $Z=-3x+4y$.
Substituting the value of E (0, 4) in Z we get,
$\begin{align}
  & Z=-3\left( 0 \right)+4\left( 4 \right) \\
 & \Rightarrow Z=16 \\
\end{align}$
Substituting the value of B (2, 3) in Z we get,
$\begin{align}
  & Z=-3\left( 2 \right)+4\left( 3 \right) \\
 & \Rightarrow Z=-6+12=6 \\
\end{align}$
Substituting the value of C (4, 0) in Z we get,
$\begin{align}
  & Z=-3\left( 4 \right)+4\left( 0 \right) \\
 & \Rightarrow Z=-12+0=-12 \\
\end{align}$
Substituting the value of D (0, 0) in Z we get,
$\begin{align}
  & Z=-3\left( 0 \right)+4\left( 0 \right) \\
 & \Rightarrow Z= 0+0= 0 \\
\end{align}$
From the above substitution of E, B, C, and D in Z we have found that when substituting the point E in Z we have got the maximum value of Z i.e. 16.
Hence, Z is maximized at point E (2, 3) with the value of 16.

Note: You can mark the points E and B quite easily on the graph but to locate the point B on the graph is difficult and you might mark the wrong point on the graph paper so it is better to find the point B by the intersection of the two equations $x+2y=8\And 3x+2y=12$.
We are going to find the intersection point of the two lines i.e.
$\begin{align}
  & x+2y=8........Eq.(1) \\
 & 3x+2y=12......Eq.(2) \\
\end{align}$
Subtracting eq. (1) from eq. (2) we get,
$\begin{align}
  & 3x+2y=12 \\
 & \dfrac{-x+2y=8}{2x=4} \\
\end{align}$
Simplifying the above equation we get,
$\begin{align}
  & 2x=4 \\
 & \Rightarrow x=2 \\
\end{align}$
Plugging this value of x in eq. (2) we get,
$\begin{align}
  & x+2y=8 \\
 & \Rightarrow 2+2y=8 \\
\end{align}$
$\begin{align}
  & \Rightarrow 2y=6 \\
 & \Rightarrow y=3 \\
\end{align}$
Hence, from the above calculations we have got the value of point B (2, 3).