Question

Formulate the linear program that maximizes their profit.

Imagine you work for a company that manufactures phones, tablets and watches. Supply chain disruptions have resulted in a limited supply of three components. The company has access to 2000 units of smart glass at a price of $20 per unit. They have access to (350 + n) cameras at a price of $30 per camera. They have access to 100 dials at a price of $10 per dial. The company charges $400 per phone. A phone needs 4 units of smart glass, 3 cameras and $150 of other components. They charge $800 per tablet. A tablet needs 16 units of smart glass, 2 cameras and $250 of other components. They charge $200 per watch. A watch needs 1 unit of smart glass, 1 camera, 1 dial and $20 of other components.

Answer 1

solution : Given info of company that manufactures phones, tablets and watches. Smart glasses = 2000 units, price = $ 20/unit cameras = ( 350+n), price = $ 30 percamera diale = 100, price = $10/dial. Charge Perphone = $100 Phone needs 4 units of Smart glass 3 Cameras $150 of other components. tablet cost = $ poo tablet needs Ibunits of smart gloss & Cameras $250 of other components. watch cost = $200 watch needs = 1 unit of smart glass I camera Idial 820 other components

· linear program that maximise their profit ;8; let x, be no of phones manufactured a be noof tablets manufactured Xg be noof watches manufactured To maximise the profit, maximise Z-84004, +8800X2+8200 X3 Subject to the following constraints 4X1+16X2 + x3 = 2000 3x1+242 +83 < 350th 15x1 + 25x₂ + 3X3 5 100. - o-