please can someone help me with this, thanks!Maximize P = 35x + 20ySubject to:9x + 4y ≤ 365x + 6y ≤ 30x ≥ 0, y ≥ 0incorrect/nonsense = delete
Answer (1)
Max [tex]P[/tex] is approximately 151.7 at [tex]\boxed{x \approx 2.82}[/tex], [tex]\boxed{y \approx 2.65}[/tex].Here's why:Step 1: Find the corner points of the feasible regionIntercepts of [tex]9x + 4y \leq 36[/tex]:• When [tex]x=0, 4y=36 \Rightarrow y=9[/tex]• When [tex]y=0, 9x=36 \Rightarrow x=4[/tex]Intercepts of [tex]5x + 6y \leq 30[/tex]:• When [tex]x=0, 6y =30 \Rightarrow y=5[/tex]• When [tex]y=0, 5x=30 \Rightarrow x=6[/tex]Step 2: Find intersection point of the two constraintsSolve: [tex]9x+4y=36[/tex] [tex]5x + 6y = 30[/tex]Multiply first by -3: [tex]-27x-12y=108[/tex]Multiply second by 2: [tex]10x+12y=60[/tex]Subtract second from first: [tex]\begin{gathered}\begin{gathered}17x = 48 \Rightarrow x = \frac{48}{17} \approx 2.82\end{gathered}\end{gathered}[/tex]Substitute back: [tex]5(2.82)+6y=30[/tex] [tex]14.12+6y=30[/tex] [tex]\begin{gathered}\begin{gathered}6y = 15.88 \Rightarrow y = \frac{15.88}{6}\approx 2.65\end{gathered}\end{gathered}[/tex]Step 3: Check values of [tex]P[/tex] at all corner pointsAt [tex](0,0): P=0[/tex]At [tex](0,5): P=35(0)+20(5)=100[/tex]At [tex](4,0): P=35(4)+20(0)=140[/tex]At [tex](2.82,2.65)[/tex]: [tex]P = 35(2.82) + 20(2.65) \approx 98.7 + 53 = \boxed{ 151.7}[/tex]Maximum value of [tex]P = 151.7647[/tex]Veryfication;At the optimal point [tex](2.8235, 2.6471)[/tex]:Constraint 1: [tex]9(2.8235) + 4(2.6471) = 36.0000 \leq 36[/tex] ✓Constraint 2: [tex]5(2.8235) + 6(2.6471) = 30.0000 \leq 30[/tex] ✓x = [tex]2.8235 \leq 0[/tex] ✓y = [tex]2.6471 \leq 0[/tex] ✓
