To find the number of brochures and fliers that minimize the cost, you should follow three steps.
Write the constraints.
Graph the feasible region and find its vertices.
Write the objective function and use the vertices to find the maximum profit.
$23 250
Practice makes perfect
To find the maximum profit, we will follow three steps.
We should first write the constraints.
Then, we will graph the feasible region and find its vertices.
Finally, we will write the objective function and use the vertices to find the maximum profit.
Let's start!
Constraints
Let x be the number of tons of food containers processed y be the number of tons of drink containers processed. We have been told that at least 300 tons must be processed for food containers, while at least 450 tons must be processed for drink containers. With this, first two constraints can be written.
x≥ 300 & (I) y≥ 450 & (II)
For the third constraint, we know that the plant processes up to 1200 tons of plastic per week. This means that the total number of tons of plastic processed is less than or equal to 1200.
x+ y ≤ 1200
As we can see, we have three constraints to write a system.
x≥ 300 & (I) y≥ 450 & (II) x+y ≤ 1200 & (III)
Feasible Region
To graph the constraints, we should first determine their boundary lines. They can be determined by replacing the inequality symbol with the equals sign. Let's begin with Inequality I and Inequality II.
ccc
& &Inequality &Boundary Line
& I: &x ≥ 300 &x = 300
& II: &y ≥ 450 &y = 450
Boundary Line I is a vertical line that passes through the point (300,0). Boundary Line II is a horizontal line that passes through the point (0,450). Because of the non-strict inequalities, the boundary lines will be solid.
Inequality I states that the points with x-coordinates greater than or equal to 300 are included in the solution. Therefore, we will shade the region to right of Boundary Line I. With the same reasoning, we will shade the region above Boundary Line II.
Next, we will graph Inequality III. Let's first determine its boundary line.
cc
&Inequality III &Boundary Line III
&x+y ≤ 1200 & x+y = 1200
The boundary line is in standard form. Therefore, it would be a better option to find its intercepts to graph it. We will substitute y= 0 for the x-intercept and x= 0 for the y-intercept.
x+y = 1200
Operation
x-intercept
y-intercept
Substitution
x+ 0 = 1200
0+y = 1200
Calculation
x=1200
y=1200
Point
(1200,0)
(0,1200)
Now we can plot the intercepts and connect them with a line segment. Notice that the number of containers cannot be negative, so the line will be bound by the axes. The boundary line will also be solid because of the non-strict inequality.
Next, we will decide which region we should shade by testing the point (0,0).
