A chain saw requires 6 hours of assembly and a wood chipper 4 hours. A maximum of 48 hours of assembly time is available. The profit is $ 190 on a chain saw and $ 220 on a chipper. How many of each should be assembled for maximum profit?
To attain the maximum profit, assemble___ chain saws and ___ wood chippers.
To determine the number of chain saws and wood chippers to assemble for maximum profit, we can use a technique called linear programming.
Let's assign variables to represent the number of chain saws and wood chippers to be assembled. Let's say x represents the number of chain saws and y represents the number of wood chippers.
Now, we need to set up the objective function, which is the equation that represents the profit. The profit for the chain saw is $190, and the profit for the wood chipper is $220. So, the objective function is:
Profit = 190x + 220y
Next, we need to set up the constraints, which are the limitations on the assembly time. The assembly time for a chain saw is 6 hours, and the assembly time for a wood chipper is 4 hours. The total available assembly time is 48 hours. So, the constraints are:
6x + 4y ≤ 48 (assembly time constraint)
Also, the number of chain saws and wood chippers cannot be negative, so x ≥ 0 and y ≥ 0.
Now, we have the objective function and constraints set up. We can solve this linear programming problem using various methods, such as the graphical method or the simplex method. Here, we will solve it using the graphical method.
To solve graphically, we'll plot the constraints on a graph and find the feasible region where all constraints are satisfied. Then, we'll evaluate the objective function at the corner points of the feasible region to find the maximum profit.
Let's plot the assembly time constraint:
6x + 4y ≤ 48
First, rewrite the inequality as an equation:
6x + 4y = 48
To plot the line, we can choose two values of x and solve for y. For example, let's choose x = 0:
6(0) + 4y = 48
4y = 48
y = 12
So, one point on the line is (0, 12).
Now, let's choose y = 0:
6x + 4(0) = 48
6x = 48
x = 8
Another point on the line is (8, 0).
Plot these two points and draw the line. It should pass through (0, 12) and (8, 0).
Next, we need to determine the feasible region. Since the number of chain saws and wood chippers cannot be negative, the feasible region will be in the first quadrant.
Once we have the feasible region, we evaluate the objective function at the corner points to find the maximum profit. The corner points are the vertices of the feasible region.
In this case, the feasible region is a triangle with vertices at (0, 0), (0, 12), and (8, 0). The corner points are (0, 0), (0, 12), and (8, 0).
Now, substitute these coordinates into the objective function:
Profit = 190x + 220y
For (0, 0):
Profit = 190(0) + 220(0) = 0
For (0, 12):
Profit = 190(0) + 220(12) = 2640
For (8, 0):
Profit = 190(8) + 220(0) = 1520
The maximum profit is $2640, achieved when you assemble 0 chain saws and 12 wood chippers.
profit = 190 s + 220 c
hr = 6 s + 4 c = 48
so 4 c = 48 - 6s
c = 12 - 1.5 s
profit = 190 s + 220 (12 -1.5 s)
profit = 2640 +190 s -330 s = 2640 - 140 s
max when s = 0, make no saws
This makes sense because the chippers make more profit on each and each costs less. 4 c = 48 so make 12 chippers
Are you missing some other constraint?