question\na company manufactures two types of cell phones, a basic model and a pro model. the basic model…

question\na company manufactures two types of cell phones, a basic model and a pro model. the basic model generates a profit of $100 per phone and the pro model has a profit of $150 per phone. on the assembly line the basic phone requires 7 hours, while the pro model takes 11 hours. the basic phone requires 1 hour and the pro phone needs 3 hours for finishing, which includes loading software. both phones require 1 hour for testing. on a particular production run the company has available 1,540 work hours on the assembly line, 360 work hours for finishing, and 200 work hours in the testing department. what is that maximum profit from manufacturing both phones? round your answer to the nearest whole number. do not include a dollar sign or comma in your answer.\nsorry, thats incorrect. try again?

question\na company manufactures two types of cell phones, a basic model and a pro model. the basic model generates a profit of $100 per phone and the pro model has a profit of $150 per phone. on the assembly line the basic phone requires 7 hours, while the pro model takes 11 hours. the basic phone requires 1 hour and the pro phone needs 3 hours for finishing, which includes loading software. both phones require 1 hour for testing. on a particular production run the company has available 1,540 work hours on the assembly line, 360 work hours for finishing, and 200 work hours in the testing department. what is that maximum profit from manufacturing both phones? round your answer to the nearest whole number. do not include a dollar sign or comma in your answer.\nsorry, thats incorrect. try again?

Answer

Explanation:

Step1: Define variables

Let $x$ be the number of basic - model cell phones and $y$ be the number of pro - model cell phones.

Step2: Set up constraints

Assembly line constraint: $7x + 11y\leq1540$. Finishing constraint: $x + 3y\leq360$. Testing constraint: $x + y\leq200$. Also, $x\geq0,y\geq0$.

Step3: Set up the objective function

The profit function $P = 100x+150y$.

Step4: Solve the system of inequalities

First, consider the intersection of $x + y=200$ and $x + 3y=360$. Subtract the first equation from the second: $(x + 3y)-(x + y)=360 - 200$. $2y=160$, so $y = 80$. Then $x=200 - 80=120$. Next, consider the intersection of $x + y=200$ and $7x + 11y=1540$. From $x=200 - y$, substitute into $7x + 11y=1540$: $7(200 - y)+11y=1540$. $1400-7y + 11y=1540$, $4y=140$, $y = 35$, $x=200 - 35=165$. Then, consider the intersection of $x + 3y=360$ and $7x + 11y=1540$. From $x=360 - 3y$, substitute into $7x + 11y=1540$: $7(360 - 3y)+11y=1540$. $2520-21y+11y=1540$, $- 10y=-980$, $y = 98$, $x=360-3\times98=360 - 294 = 66$.

Step5: Evaluate the profit function at the corner - points

Corner - point 1: $(0,0)$, $P(0,0)=100\times0 + 150\times0=0$. Corner - point 2: Let $x = 0$ in $x + 3y=360$, then $y = 120$, $P(0,120)=100\times0+150\times120 = 18000$. Corner - point 3: Let $y = 0$ in $7x + 11y=1540$, then $x = 220$, but $x + y\leq200$, so this is not a valid corner - point. Corner - point 4: $(66,98)$ $P(66,98)=100\times66+150\times98=6600 + 14700=21300$. Corner - point 5: $(120,80)$ $P(120,80)=100\times120+150\times80=12000 + 12000=24000$. Corner - point 6: $(165,35)$ $P(165,35)=100\times165+150\times35=16500+5250 = 21750$.

Answer:

24000