Afurniture company manufactures desks and chairs. each desk uses four units of wood, and each chair uses three units of wood. a desk contributes $400 to profit, a chair contributes $250. marketing restrictions require that the number of chairs produced be at least twice the number of desks produced. there are 2000 units of wood available. maximize the profit from manufacturing. set up the algebraic model. include the three components: (decision variables, objective function and constraints) what are the decision variables, what is the objective function, and what are the constraints? write up your algebraic model.

$180000

Step-by-step explanation:

Let's c be the number of chair and d be the number of desks.

The constraint functions:

- Unit of wood available 4d + 3c <= 2000 or d <= 500 - 0.75c

- Number of chairs being at least twice of desks c >= 2d or d <= 0.5c

c >= 0

d >= 0

The objective function is to maximize the profit function

P (c,d) = 400d + 250c

We draw the 2 constraint functions (500 - 0.75c and 0.5c) on a c-d coordinates (witch c being the horizontal axis and d being the vertical axis)  and find the intersection point 0.5c = 500 - 0.75c

1.25c = 500

c = 400 and d = 0.5c = 200 so P(400, 200) = $250*400 + $400*200 = $180,000

The 500 - 0.75c intersect with c-axis at d = 0 and c = 500 / 0.75 = 666 and P(666,0) = 666*250 = $166,500

So based on the available zones in the chart we can conclude that the maximum profit we can get is $180000

-12+10=-2 so just remember your -- =+ and your -+= - if first numbers appear to be lower.

Step-by-step explanation: