A cargo ship is loaded with two types of cargo. Each container of cargo A is 400 cubic feet involume, weighs 10,000 lbs and earns $11,000 in profit. Each container of cargo B is 300 cubic feetin volume, weighs 20,000 lbs, and earns $9,000 in profit. The ship can carry no more than 30,000cubic feet in volume and no more than 1,000,000 lbs. How many containers of each type shouldbe carried to maximize the profit?Let x be the number of containers of cargo A and y be the number of containers of cargoLet's finally answer the question. How many containers of each type should be carried tomaximize the profit? Fill in the blanks.The ship should carrycontainers of cargo A andcontainers of cargo B for a maximum profit ofdollars.

The following data is given, which is shown in a table: CARGO Volume Weights A 400 cubit feet 10,000 lbs B 300 cubic feet 20,000 lbs One has to determine the number of containers of cargo A and B to maximize the profit. Given: x = number of containers of cargo A, y = number of containers of cargo B (x and y are greater than equal to zero.) Since a Ship loaded with cargo A and cargo B earns $11,000 and $9000 profit respectively, it can be expressed as follows: Profit(z) = 11000x +9000y Since the volume of both types of the container having cargo A and B can not exceed more than 30,000 cubic feet, it can be expressed as: 400x + 300y ≤ 30000 4x + 3y ≤ 300 Since the weights of both types of the container having cargo A and B can not exceed more than 1,000,000 lbs, it can be expressed as: 10000x + 20000y≤ 1000000 x + 2y ≤ 100 Thus, the problems change into the following linear programming problem: Profit(z) = 11000x +9000y 4x + 3y ≤ 300 x + 2y ≤ 100 x and y are non-negative. Sketch the graph of both inequalities in the XY-plane. The figure will be as follows; From the figure, we obtained the corner points (75, 0), (0, 50), and (60, 20) that give us maximize profit. At Point (75, 0): Profit z=1100075 +90000=825000 At point (0, 50): Profit z=110000 +900050=450000 At point (60, 20): Profit z=1100060 +900020=660000+180000=840000 Thus, it is clear that profit can be maximized if x = 60 and y = 20. The ship should carry 60 containers of cargo A and 20 containers of cargo B for a maximum profit of 840000 dollars.