a contractor builds two models of homes, model a and model b. model a requires 400 worker - days of labor…

a contractor builds two models of homes, model a and model b. model a requires 400 worker - days of labor and $400,000 in capital, and model b requires 500 worker - days of labor and $250,000 in capital. the contractor has a total of 10,000 worker - days and $6,000,000 in capital available per month. the profit is $120,000 on model a and $75,000 on model b. building how many of each model will maximize the monthly profit? what is the maximum possible profit? building how many of each model will maximize the monthly profit? profit is maximized with 15 models a homes and model b homes, with model a homes and 16 model b homes, or with 10 model a homes and model b homes.

a contractor builds two models of homes, model a and model b. model a requires 400 worker - days of labor and $400,000 in capital, and model b requires 500 worker - days of labor and $250,000 in capital. the contractor has a total of 10,000 worker - days and $6,000,000 in capital available per month. the profit is $120,000 on model a and $75,000 on model b. building how many of each model will maximize the monthly profit? what is the maximum possible profit? building how many of each model will maximize the monthly profit? profit is maximized with 15 models a homes and model b homes, with model a homes and 16 model b homes, or with 10 model a homes and model b homes.

Answer

Explanation:

Step1: Define variables

Let $x$ be the number of model A homes and $y$ be the number of model B homes.

Step2: Set up constraints

Labor - constraint: $400x + 500y\leq10000$, which simplifies to $4x + 5y\leq100$. Capital - constraint: $400000x+250000y\leq6000000$, which simplifies to $8x + 5y\leq120$. Also, $x\geq0,y\geq0$ and $x,y$ are non - negative integers.

Step3: Express the profit function

The profit function $P = 120000x+75000y$.

Step4: Solve the system of inequalities for corner points

From $4x + 5y=100$, when $x = 0$, $y = 20$; when $y = 0$, $x = 25$. From $8x + 5y=120$, when $x = 0$, $y = 24$; when $y = 0$, $x = 15$. Solve the system $\begin{cases}4x + 5y=100\8x + 5y=120\end{cases}$ Subtract the first equation from the second: $(8x + 5y)-(4x + 5y)=120 - 100$, $4x=20$, $x = 5$, then substitute $x = 5$ into $4x+5y=100$, $20 + 5y=100$, $y = 16$. The corner points of the feasible region are $(0,0),(0,20),(15,0),(5,16)$.

Step5: Evaluate the profit function at corner points

At $(0,0)$: $P(0,0)=120000\times0 + 75000\times0=0$. At $(0,20)$: $P(0,20)=120000\times0+75000\times20 = 1500000$. At $(15,0)$: $P(15,0)=120000\times15+75000\times0=1800000$. At $(5,16)$: $P(5,16)=120000\times5+75000\times16=600000 + 1200000=1800000$.

Answer:

Profit is maximized with 15 model A homes and 0 model B homes, with 5 model A homes and 16 model B homes, or with 10 model A homes and 12 model B homes. (To find the last one: From $4x + 5y=100$, when $x = 10$, $y = 12$. And $P(10,12)=120000\times10+75000\times12=1200000 + 900000=2100000$ which is also a candidate for maximum). The maximum profit is $$2100000$. So the blanks are filled as: 0, 5, 12.