national business machines manufactures x model a portable printers and y model b portable printers. each…

national business machines manufactures x model a portable printers and y model b portable printers. each model a costs $100 to make, and each model b costs $150. the profits are $45 for each model a and $50 for each model b portable printer. if the total number of portable printers demanded per month does not exceed 2500 and the company has earmarked no more than $600,000/month for manufacturing costs, how many units of each model should national make each month to maximize its monthly profit? (x, y) = ( ) what is the optimal profit? $

national business machines manufactures x model a portable printers and y model b portable printers. each model a costs $100 to make, and each model b costs $150. the profits are $45 for each model a and $50 for each model b portable printer. if the total number of portable printers demanded per month does not exceed 2500 and the company has earmarked no more than $600,000/month for manufacturing costs, how many units of each model should national make each month to maximize its monthly profit? (x, y) = ( ) what is the optimal profit? $

Answer

Explanation:

Step1: Set up constraints

Cost constraint: $100x + 150y\leq600000$, simplifies to $2x + 3y\leq12000$. Quantity - constraint: $x + y\leq2500$, and $x\geq0,y\geq0$.

Step2: Set up profit function

The profit function $P = 45x+50y$.

Step3: Solve the system of inequalities

We can rewrite $x + y\leq2500$ as $y\leq - x + 2500$ and $2x + 3y\leq12000$ as $y\leq-\frac{2}{3}x + 4000$. The intersection of $x + y=2500$ and $2x + 3y=12000$: Substitute $y = 2500 - x$ into $2x + 3y=12000$, we get $2x+3(2500 - x)=12000$. $2x + 7500-3x=12000$, $-x=12000 - 7500$, $x=-4500$ (not valid). Intersection of $y = 0$ and $2x+3y = 12000$ gives $x = 6000$ (but violates $x + y\leq2500$). Intersection of $y = 0$ and $x + y=2500$ gives $(x,y)=(2500,0)$. Intersection of $x = 0$ and $x + y=2500$ gives $(x,y)=(0,2500)$. Intersection of $x = 0$ and $2x + 3y=12000$ gives $(x,y)=(0,4000)$ (violates $x + y\leq2500$). Evaluate the profit function $P = 45x + 50y$ at the corner - points: At $(2500,0)$: $P=45\times2500+50\times0 = 112500$. At $(0,2500)$: $P=45\times0+50\times2500=125000$.

Step4: Determine the optimal solution

The maximum profit occurs when $(x,y)=(0,2500)$. The optimal profit $P = 50\times2500=125000$.

Answer:

$(x,y)=(0,2500)$ $125000$