Answer:
a. See below.
b. See attachment.
c. (1, 2) (2, 0) (0, 0) (0, 1)
d. The maximum value of P is 11.
Step-by-step explanation:
[tex]\textsf{Maximize}: \quad P=-x+6y[/tex]
[tex]\textsf{Subject to}:\begin{cases}\begin{aligned} \quad y &\leq -2x+4\\y&\leq x+1\\x&\geq 0\\y & \geq 0\end{aligned}\end{cases}[/tex]
Graph the lines:
[tex]\textsf{Draw the line } \;\;y=-2x+4 \;\;\textsf{and shade under the line}.[/tex]
[tex]\textsf{Draw the line } \;\;y=x+1 \;\;\textsf{and shade under the line}.[/tex]
[tex]\textsf{Draw the line } \;\;x= 0 \;\;\textsf{and shade above the line}.[/tex]
[tex]\textsf{Draw the line } \;\;y=0\;\;\textsf{and shade above (to the right of) the line}.[/tex]
Therefore, the feasible region is bounded by the vertices:
- A = (1, 2)
- B = (2, 0)
- C = (0, 0)
- D = (0, 1)
Determine the value of P at the vertices by substituting the x and y values of the points into the equation for P:
[tex]\textsf{Value of $P$ at $A(1, 2)$}: \quad -(1)+6(2)=11[/tex]
[tex]\textsf{Value of $P$ at $B(2, 0)$}: \quad -(2)+6(0)=-2[/tex]
[tex]\textsf{Value of $P$ at $C(0, 0)$}: \quad -(0)+6(0)=0[/tex]
[tex]\textsf{Value of $P$ at $D(0, 1)$}: \quad -(0)+6(1)=6[/tex]
Hence, the maximum value of P is 11 at vertex (1, 2).
