Question 1118425: A manufacturer produces two models of mountain bikes. Model A requires 5 hours of assembly time and 2 hours of painting time, and Model B requires 4 hours of assembly time and 3 hours of painting time. The maximum total weekly hours available in the assembly department and the painting department are 200 hours and 108 hours, respectively. The profits per unit are $25 for Model A and $15 for Model B. How many of each type should be produced to maximize profit?
___ bikes of Model A
___ bikes of Model B
What is the maximum profit? $____
Found 2 solutions by greenestamps, ikleyn:
Answer by greenestamps(13200)   (Show Source):
You can put this solution on YOUR website!
Let x be the number of model A produced and y be the number of model B. Then the equation of the boundary line for the assembly time inequality is
and the equation for boundary line for the painting time is
The standard procedure for solving the problem would be this:
(1) Graph the equations of the two boundary lines and find where they intersect;
(2) Use the inequalities to determine the corners of the feasibility region; and
(3) Evaluate the objective function (the profit function 25x+15y) at each corner.
I leave the details to you. The corners of the feasibility region are (0,0), (0,36), (24,20), and (40,0). The profits at the respective corners are 0, 540, 900, and 1000. So the maximum profit is when the manufacturer produces 40 of model A and 0 of model B.
Not a very reasonable scenario....
But for this example you don't need to do all that work.
In fact, in the standard process for solving this kind of problem, you do NOT need to find the value of the objective function at each corner of the feasibility region. You can tell which corner will yield the maximum value of the objective function by comparing the slope of the objective function to the slopes of the constraint lines.
In a reasonable problem with two constraints, the slope of the objective function will be between the slopes of the two constraint boundary line functions; that means the maximum value of the objective function will be at the intersection of the constraint boundary line.
But if the slope of the objective function is smaller (more negative) or larger (less negative) than the slope of either constraint boundary line function, then the maximum value of the objective function will be at the x- or y-intercept of one of the constraint boundary line functions.
In your example, the slopes of the constraint boundary line functions are -5/4 and -2/3; the slope of the objective function is -5/3. -5/3 is more negative than -5/4, so the maximum profit will NOT be at the intersection of the constraint function boundary lines.
So here is all that is required to find the solution to your problem.
(1) Graph the equations of the two boundary lines to find two corners of the feasibility region -- (40,0) and (0,36);
(2) Find the slopes of the two constraint boundary lines and of the objective function and observe that the slope of the objective function is more negative than the slope of either constraint boundary line;
(3) immediately know that the maximum profit will be at (40,0); and
(4) evaluate the objective function at (40,0) to find the maximum profit of $1000.
This shortcut of comparing slopes of the boundary lines and of the objective function saves a bit of time in a problem where there are two constraints, because there is no need to find the point of intersection of the boundary lines.
It can save a LOT of time if there are three or more constraint boundary lines, because the slopes will tell you immediately which intersection point will give the maximum value of the objective function; so you will only need to find the coordinates of one of the many intersection points.
Answer by ikleyn(52794)  (Show Source):
You can put this solution on YOUR website! .
A manufacturer produces two models of mountain bikes.
Model A requires 5 hours of assembly time and 2 hours of painting time,
and Model B requires 4 hours of assembly time and 3 hours of painting time.
The maximum total weekly hours available in the assembly department and the painting department are
200 hours and 108 hours, respectively.
The profits per unit are $25 for Model A and $15 for Model B. How many of each type should be produced to maximize profit?
~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~
Let X be the number of bikes Model A;
Y be the number of bikes Model B.
Then the constraints are
5X + 4Y <= 200, (1) (Assembly time constraint of 200 hours)
2X + 3Y <= 108. (2) (Painting time constraint of 108 hours)
X >= 0, Y >= 0. (3) (Non-negativity constraint)
The profit function is
P(X,Y) = 25X + 15Y. (4)
The feasibility domain is shown in the Figure below.
It is the quadrilateral in the Quadrant I below two straight lines that are constraints.
Plots 5x + 4y = 200 (red line) and 2x + 3y = 108 (green line).
The vertices of the quadrilateral are
P1 = (0,36) (y-intersept of the green line)
P2 = (40,0) (x-intercept of the red line)
P3 = (24,20) (the intersection point of the red and green lines) <<<---=== You may find it by solving the system of equations
P4 = (0,0) (the origin of the coordinate system)
According to the Linear Programming method, you need to find the values of the profit function in the vertices of the feasibility domain,
and then select the vertex, where the value of the profit function is maximal.
These calculations re shown below:
at P1: P(X,Y) = P(0,36) = 0*25 + 36*15 = 540;
at P2: P(X,Y) = P(40,0) = 40*25 + 0*15 = 1000;
at P3: P(X,Y) = P(24,20) = 24*25 + 20*15 = 900;
at P4: P(X,Y) = P(0,0) = 0*25 + 0*15 = 0.
The maximum value is 1000 at P2; so, X = 40, Y = 0 is the solution to the given problem.
Answer. 40 bikes of the model A and 0 (zero, ZERO) bikes of the model B satisfy the constraints
and provide the maximum profit of 1000 dollars.
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