The area of a parking lot is 1710 square meters. A car requires 5 square meters and a bus requires 32 square meters of space. There can be at most 180 vehicles parked at one time. If the cost to park a car is $2.00 and a bus is $6.00, how many buses should be in the lot to maximize income? Please help :(

Answer:To maximize the income should be 30 buses and 150 carsStep-by-step explanation:Letx-----> the number of carsy ----> the number of buswe know that[tex]5x+32y\leq1,710[/tex] ------> inequality A[tex]x+y\leq 180[/tex] ----> inequality BThe function of the cost to maximize is equal to[tex]C=2x+6y[/tex]Solve the system of inequalities by graphingThe solution is the shaded areasee the attached figureThe vertices of the solution are(0,0),(0,53),(150,30),(180,0)Verify(0,53) ---> [tex]C=2(0)+6(53)=\$318[/tex](150,30) ---> [tex]C=2(150)+6(30)=\$480[/tex]thereforeTo maximize the income should be 30 buses and 150 cars