By graphing the system of constraints find the values of x and y that maximize the objective function, find the maximum value.3x+y<=7x+2y<=9x>=0y>=0Maximum for P=2x+yA. P=3B. P=4C. P=6D. P=10

Answer:x=1, y=4 P=6Step-by-step explanation:We are given that [tex]3x+y\leq 7[/tex][tex]x+2y\leq 9[/tex][tex]x\geq 0[/tex][tex]y\geq 0[/tex]Objective function [tex]P=2x+y[/tex]We have to values of x and y that maximize the P and maximum value of P .First we change inequality equation into equality equation[tex]3x+y=7[/tex]  (I equation )[tex]x+2y=9[/tex]  (II equation)Equation I multiply by 2 and then subtract from II equation [tex]-5x=-5[/tex][tex]x=1[/tex]Substitute x=1 in equation IThen, we get[tex]3(1)+y=7[/tex][tex]3+y=7[/tex][tex]y=7-3=4[/tex]The two equation intersect at point (1,4).Substitute x=0 in equation IThen , we get y=7Substitute y=0 then [tex]3x=7[/tex][tex]x=\frac{7}{3}=2.3[/tex]The equation I cut the x- axis at point (2.3,0)and y-axis at (0,7).Substitute x=0 in equtaion II[tex]2y=9[/tex][tex]y=\frac{9}{2}=4.5[/tex]Substitute y=0Then, we get [tex]x=9[/tex]Therefore, the equation II cut the x- axis at point (9,0) and y axis at point (0,4.5).Substitute x=0 and y=0 in inequality Equation I [tex]3(0)+0=0< 7[/tex]It is true . Therefore, shaded region below the line.Substitute x=0 and y=0 in inequality equation IIThen, [tex]0+2(0)=0 < 9[/tex]It is true. Therefore, the shaded region below the line.The feasible region is bounded.The feasible region bounded by (0,0),(0,4.5),(2.3,0) and (1,4).At (0,4.5)[tex]P=2(0)+4.5=4.5[/tex]At (2.3,0)[tex]P=2(2.3)+0=4.6[/tex]At (1,4)[tex]P=2(1)+4=6[/tex]At (0,0)[tex]P=2(0)+0=0[/tex]Hence, maximum value of P is 6 at (1,4).x=1 and y=4