Question

A confectioner sells two types of nut mixtures. The standard-mix package and the deluxe-mix package. The

standard-mix contains 50 grams of cashews and 120 grams of peanuts and sells for $1.95. The deluxe-mix
contains 100 grams of cashews and 40 grams of peanuts and sells for $2.25. The confectioner has 12 kilograms
of cashews and 18 kilograms of peanuts available. Based on pasts sales, the confectioner needs to have at least as
many standard as deluxe packages available.
Note, if we let x represent the number of standard-mix packages and y represent the number of deluxe-mix
packages sold, then we know the revenue produced by selling the packages is given by
(, ) = 1.95 + 2.25
Clearly, to maximize revenue the confectioner would simply sell as many packages as possible. However, it is
not as simple as that since there are some “constraints” that must be satisfied.
Constraint 1: There are 12,000 grams of cashews available. Since the standard-mix uses 50 grams and the
deluxe-mix uses 100 grams, we need to ensure that
50 + 100 ≤ 12000 ⇒ + 2 ≤ 240
Constraint 2: There are 18,000 grams of peanuts available. Since the standard-mix uses 120 grams of the deluxe mix uses 40 grams, we need to ensure that
120 + 40 ≤ 18000 ⇒ 3 + ≤ 450
Constraint 3: The confectioner needs to have at least as many standard-mix packages as she has deluxe-mix
packages, so we also require that
≥ ⇒ ≤
Constraints 4 and 5: Since the confectioner cannot produce a negative number of packages, it must be that
≥ 0 and ≥ 0

If we plot the inequalities given by the constraints, the ordered pair (x, y) that maximizes the revenue function
will be a vertex of the “feasible region.” The feasible region is the region in the xy - plane defined by the
constraints and can be seen in the figure here.
Questions
1. Find the vertices of the feasible region.
a. Find the coordinates of vertex 1.
b. Find the coordinates of vertex 2.
c. Find the coordinates of vertex 3.
d. Find the coordinates of vertex 4.
2. Find the value of the revenue function R at each vertex.
a. Value of R at vertex 1.
b. Value of R at vertex 2.
c. Value of R at vertex 3.
d. Value of R at vertex 4.
3. How many standard-mix packages and how many deluxe-mix packages will the confectioner need to see
to maximize her revenue?

Answer 1

1a) Vertex 1 = (0, 0)

1b) Vertex 2 = (80, 80)

1c) Vertex 3 = (132, 54)

1d) Vertex 4 = (80, 0)

2a) R = 0

2b) R = 336

2c) R = 378.9

2d) R = 156

3) 132 standard-mix packages and 54 deluxe-mix packages.

Explanation:

A confectioner sells two types of nut mixtures.

Let x be the number of standard-mix packages sold.

Let y be the number of deluxe-mix packages sold.

The revenue produced by selling the packages is given by the equation:

`R(x, y) = 1.95x + 2.25y`

We have been given five constraints represented by this system of inequalities:

`\begin{cases}x+2y\leq 240\\3x+y\leq 450\\y \leq x\\x \geq0\\y \geq 0\end{cases}`

To maximize the revenue produced by selling the packages subject to the given constraints, we need to analyse the feasible region defined by the system of inequalities.

The feasible region for the given constraints is the region where all the shaded areas overlap. Therefore, we need to graph the system of inequalities given by the constraints.

Rearrange the first two inequalities so that they are in slope-intercept form:

`\boxed{\begin{aligned}x+2y&\leq 240\\2y&\leq-x+240\\y&\leq -(1)/(2)x+120\end{aligned}}`
`\boxed{\begin{aligned}3x+y&\leq 450\\y&\leq-3x+450\\\vphantom{\frac12}\end{aligned}}`

When graphing an inequality with the sign ≥, draw a solid line and shade above the line.

When graphing an inequality with the sign ≤, draw a solid line and shade below the line.

Graph the inequalities (see attachment):

• Draw a solid line at y = -(1/2)x + 120 and shade below the line.
• Draw a solid line at y = -3x + 450 and shade below the line.
• Draw a solid line at y = x and shade below the line.
• For x ≥ 0 shade to the right of the y-axis.
• For y ≥ 0 shade above the x-axis.

The feasible region has four corner points (vertices).

Vertex 1 is the point of intersection of y = x, the x-axis and the y-axis.

Therefore, vertex 1 is the origin (0, 0).

Vertex 2 is the point of intersection between y = x and x + 2y = 240. Substitute the first equation into the second equation and solve for x. Then substitute the found value of x into the first equation and solve for y:

`\begin{aligned}x + 2x &= 240\\3x&=240\\x&=80\end{aligned}`
`\begin{aligned}\vphantom{x + 2x &= 240}\\y&=x\\y&=80\end{aligned}`

Therefore, vertex 2 is (80, 80).

Vertex 3 is the point of intersection between x + 2y = 240 and 3x + y = 450. Rearrange the second equation to isolate y, substitute it into the first equation and solve for x. Then substitute the found value of x into the second equation and solve for y:

`\begin{aligned}x + 2(-3x+450) &= 240\\x-6x+900&=240\\-5x&=-660\\x&=132\end{aligned}`
`\begin{aligned}3(132) + y &= 450\\396+y&=450\\y&=450-396\\y&=54\end{aligned}`

Therefore, vertex 3 is (132, 54).

Vertex 4 is the point of intersection of 3x + y = 240 and the x-axis.

Substitute y = 0 into the equation and solve for x:

`\begin{aligned}3x + 0 &= 240\\3x&=240\\x&=80\end{aligned}`

Therefore, vertex 4 is (80, 0).

In summary, the four vertices are:

• Vertex 1 = (0, 0)
• Vertex 2 = (80, 80)
• Vertex 3 = (132, 54)
• Vertex 4 = (80, 0)

Determine the value of R at each vertex by substituting the x and y values of the points into the equation for R:

`\begin{aligned}&\textsf{Vertex\;1:}\;\;(0,0) \implies &R&=1.95(0) + 2.25(0)=\boxed{0}\\&\textsf{Vertex\;2:}\;\;(80, 80) \implies &R&=1.95(80) + 2.25(80)=\boxed{336}\\&\textsf{Vertex\;3:}\;\;(132, 54) \implies &R&=1.95(132) + 2.25(54)=\boxed{378.9}\\&\textsf{Vertex\;4:}\;\;(80, 0) \implies &R&=1.95(80) + 2.25(0)=\boxed{156}\\\end{aligned}`

The maximum value of R is $378.90 at vertex (132, 54).

Therefore, the confectioner needs to sell 132 standard-mix packages and 54 deluxe-mix packages to maximize her revenue.