To find the number of each cake that maximize the revenue, you will 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 revenue.
225 yellow cakes and 0 strawberry cakes
Practice makes perfect
To find the number of each cake that maximize the revenue, 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 revenue.
Let's start!
Constraints
Let x be the number of strawberry cakes and y be the number of yellow cakes. We know that the number of cakes cannot be negative. With this, first two constraints can be written.
x≥ 0 & (I) y≥ 0 & (II)
For the third constraint, we will use the given data and make an organized table to write it.
Verbal Expression
Algebraic Expression
Assembly time for x number of strawberry cakes (h)
3 x
Assembly time for y number of yellow cakes (h)
2 y
Total assembly time is less than or equal to 450 hours.
3 x+ 2 y≤ 450
As we can see, we have three constraints to write a system.
x≥ 0 & (I) y≥ 0 & (II) 3x+2y ≤ 450 & (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.
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& &Inequality &Boundary Line
& I: &x ≥ 0 &x = 0
& II: &y ≥ 0 &y = 0
Notice that the boundary lines are the axes and the inequalities refer to Quadrant I. This means that we will shade Quadrant I. Because of the non-strict inequalities, the boundary lines will be solid.
Next, we will graph Inequality III. Let's first determine its boundary line.
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&Inequality III &Boundary Line III
&3x+2y ≤ 450 &3x+2y = 450
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.
3x+2y = 450
Operation
x-intercept
y-intercept
Substitution
3x+2( 0) = 450
3( 0)+2y = 450
Calculation
x=150
y=225
Point
(150,0)
(0,225)
Now we can plot the intercepts and connect them with a line segment. Notice that the number of cakes 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).
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 intercepts of Boundary Line III and the origin are the vertices of the feasible region.
Vertices
(150,0), (0,0), (0,225)
Objective Function and Maximum Revenue
Let R be the total revenue. We will make an organized table to write a function for the total revenue.
Verbal Expression
Algebraic Expression
Revenue from x number of strawberry cakes ($)
35 x
Revenue from y number of yellow cakes ($)
25 y
Total revenue is $R.
R= 35 x+ 25 y
To find the number of each cake that maximize the revenue, we will substitute the vertices into the function.
Vertex
35x+25y
R
( 0, 0)
35( 0)+25( 0)
$0
( 150, 0)
35( 150)+25( 0)
$5250
( 0, 225)
35( 0)+25( 225)
$5625
As a result, to maximize the revenue, 225 yellow cakes and 0 strawberry cake should be made. With this, the maximum revenue will be $5625.
