a division of the winston furniture company manufactures dining - tables and chairs. each table requires 40…

a division of the winston furniture company manufactures dining - tables and chairs. each table requires 40 board - feet of wood and 3 labor - hours. each chair requires 16 board - feet of wood and 4 labor - hours. the profit for each table is $45, and the profit for each chair is $20. in a certain week, the company has 3200 board - feet of wood available and 520 labor - hours available. how many tables and chairs should winston manufacture to maximize its profit? (let x represent the number of tables winston manufactures and let y represent the number of chairs it manufactures.)\nwhat is the maximum profit?

a division of the winston furniture company manufactures dining - tables and chairs. each table requires 40 board - feet of wood and 3 labor - hours. each chair requires 16 board - feet of wood and 4 labor - hours. the profit for each table is $45, and the profit for each chair is $20. in a certain week, the company has 3200 board - feet of wood available and 520 labor - hours available. how many tables and chairs should winston manufacture to maximize its profit? (let x represent the number of tables winston manufactures and let y represent the number of chairs it manufactures.)\nwhat is the maximum profit?

Answer

Explanation:

Step1: Set up constraints

The wood - constraint: $40x + 16y\leq3200$ (since each table needs 40 board - feet and each chair needs 16 board - feet and total wood is 3200 board - feet). Simplifying gives $5x + 2y\leq400$. The labor - constraint: $3y+4x\leq520$ (since each table needs 4 labor - hours and each chair needs 3 labor - hours and total labor - hours is 520). Also, $x\geq0,y\geq0$ as the number of tables and chairs cannot be negative.

Step2: Set up the profit function

The profit function $P = 45x+20y$.

Step3: Solve the system of inequalities for corner points

From $5x + 2y=400$, we get $y = 200-\frac{5}{2}x$. From $4x + 3y=520$, we get $y=\frac{520 - 4x}{3}$. Solve the system $\begin{cases}5x + 2y=400\4x + 3y=520\end{cases}$. Multiply the first equation by 3 and the second by 2: $15x+6y = 1200$ and $8x+6y = 1040$. Subtract the second from the first: $15x - 8x=1200 - 1040$, $7x = 160$, $x=\frac{160}{7}\approx22.86$. Then substitute $x$ into $y = 200-\frac{5}{2}x$ or $y=\frac{520 - 4x}{3}$. The corner - points of the feasible region are found by considering the intersection of the boundary lines and the non - negativity constraints: Intersection of $x = 0$ and $5x+2y = 400$ gives $(0,200)$; $P(0,200)=4000$. Intersection of $y = 0$ and $5x+2y = 400$ gives $(80,0)$; $P(80,0)=3600$. Solve the system of $5x + 2y=400$ and $4x + 3y=520$: From $5x+2y = 400$, we have $y = 200-\frac{5}{2}x$. Substitute into $4x + 3y=520$: $4x+3(200-\frac{5}{2}x)=520$. $4x + 600-\frac{15}{2}x=520$. $8x+1200 - 15x = 1040$. $- 7x=-160$, $x=\frac{160}{7}\approx22.86$, $y=\frac{520-4\times\frac{160}{7}}{3}=\frac{520\times7 - 640}{21}=\frac{3640 - 640}{21}=\frac{3000}{21}\approx142.86$. The intersection of $x = 0$ and $4x + 3y=520$ gives $(0,\frac{520}{3}\approx173.33)$; $P(0,\frac{520}{3})=\frac{10400}{3}\approx3466.67$. The intersection of $y = 0$ and $4x + 3y=520$ gives $(130,0)$; $P(130,0)=5850$. After substituting and calculating, we find that when $x = 80$ and $y = 0$, the profit is maximized.

Answer:

$(80,0)$ $3600$