Could someone please check my work

Solve the following linear programming problem using The 6) algebraic method. insert sleek variables Maximize 2 = 3 x+4y 1 into line equations Subject 2x+4 54 2xty + p = 4 TO - x+24 + 9 = 4 Constraints - x+ 24 54 x, 4 20 (4 ) = F - " * , 4 , p 1 9 20 " 4 ! = 6 cases (4- 2) ! 2! " subject to equations " extreme points / ( x ,y) = (0, 0) = 2(0 ) + 0 + p = 4 => p= 4 = ( x , y ) = 10,0) = feasible - (0)+2(0) + 9=4 =9-4 intersection Point ( x, p ) = 10, 0) 7 2 ( 0 ) + y to = 4 = 4= 4 - (0) +24 + q=4 = 2(4 ) +9=4 -79=-4 Lo negative / doesn't Satisfy constraints So ( 4,4 ) = (0, 4 ) = ( x, q) = (0, 0) = 2(0) +4 + p= 4 37 2 +p= 4= ) p= 2 infeasible inter - Section point - (0) + 24 +0-4 = 24=4-74 = 2 7 ( 4 , 4 ) = 10, 2 ) = feasible intersection ( 4, p ) = (0, 0) =7 2*+0 +0=4= *= 2 - * + 260) +9 # 24 27 - 2 +9=4 =7 ( x , y ) = ( 2, 0 ) = "-X+ 2(0) + 0= 4 7 * = - 4 ( 4,9) =60,0) = 2 x +0 +0 = 4 2 #= ( x,4 ) = ( - 4, 0 ) = feasible htersection ( P. 9) = 60, 0) 7 2 * +y +0 = 4 5/ 2 xtyzy - x+ 24 to = 4 -> ( x, 4 ) = 10.8, ( - X + 2y = 4 ) - 2 =davy = feasible 2x - 4y =-8 ( 2 x + 4 = 4 ) intersection paint U - 5y =-12 4 = 12 : -2.4 5 = > 2 x + = = 4 = ) x + 2-2 => x= 0.8 4 . Extreme Points : (0, 0) , (0, 2), (2, 0), (0.8, 2.4) For (0 , 0 ) : 2 = 3(0 ) + 4 (0 ) = 0 For (0, 2 ) : 2 = 3(0 ) +4(2 ) = 8 For ( 2 , 0 ) : 2 = 3 (2 ) + 410 ) = 6 For (0.8, 2.4 ) : 2 = 3 (0.8 ) +4( 2.4 ) = 12 - maximized solution verified using Excel's " Solver" add - in