A confectioner sells two types of nut mixtures. The standard-mixture package contains 100 g of cashews and 200 g of peanuts and sells for $1.85. The deluxe-mixture package contains 150 g of cashews and 50 g of peanuts and sells for $2.45. The confectioner has 15 kg of cashews and 20 kg of peanuts available. On the basis of past sales, the confectioner needs to have at least as many standard as deluxe packages available. How many bags of each mixture should he package to maximize his revenue?

Answer 1

Let $x$ be the number of standard-mixture packages and $y$ be the number of deluxe-mixture packages.

From the problem description we have the following constraints:

Cashews:
$100x + 150y \leq 15000$ (1)

Peanuts:
$200x + 50y \leq 20000$ (2)

Number of standard packages should be at least as many as deluxe packages:
$x \geq y$

The revenue to be maximized is given by:
$R = 1.85x + 2.45y$ (3)

We will now solve the constraints:

1) $100x + 150y \leq 15000$

Dividing both sides by $50$ we get:

$2x + 3y \leq 300$

2) $200x + 50y \leq 20000$

Dividing both sides by $50$ we get:

$4x + y \leq 400$

Now we plot the inequality constraints graphically and find the boundary point on which the function $R(x, y) = 1.85x + 2.45y$ is maximized.

The inequality constraints plot looks like the following:

y
^
|
400|
|
+-------x.

and the relevent boundary points are $A(100, 100), \ B(150, 0)$.

Now we plug in these boundary points into the revenue function.

$R_A(100, 100) = 1.85 \times 100 + 2.45 \times 100 = 430 $

$R_B(150, 0) = 1.85 \times 150 + 2.45 \times 0 = 277.5 $

Since the revenue at point A is higher, the confectioner should package 100 standard-mixture packages and 100 deluxe-mixture packages to maximize his revenue.