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To find the number of brochures and fliers that minimize the cost, you should follow three steps.
$23 250
To find the maximum profit, we will follow three steps.
Let's start!
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.
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).
x= 0, y= 0
Zero Property of Multiplication
Since the point satisfies the inequality, region that contains the point will be shaded. Let's do it!
The overlapping section of the graph above represents the feasible region. The points of intersection are the vertices of the feasible region.
Vertices (300,450), (300,900), (750,450)
Let P be the total cost. We will make an organized table to write a function for the total profit.
| Verbal Expression | Algebraic Expression |
|---|---|
| Profit on x tons of food containers ($) | 17.5 x |
| Profit on y tons of drink containers ($) | 20 y |
| Total profit is $P. | P= 17.5 x+ 20 y |
To find the number of tons of each container that maximize the profit, we will substitute the vertices into the function.
| Vertex | 17.5x+20y | P |
|---|---|---|
| ( 300, 450) | 17.5( 300)+20( 450) | $14 250 |
| ( 300, 900) | 17.5( 300)+20( 900) | $22 125 |
| ( 750, 450) | 17.5( 750)+20( 450) | $23 250 |
As a result, the maximum profit is $23 250.