Question

A candy manufacturer has 130 pounds of chocolate-covered cherries and 170 pounds of chocolate-covered mints in stock. He decides to sell them in the form of two different mixtures. One mixture will contain half cherries and half mints by weight and will sell for $2.00 per pound. The other mixture will contain one-third cherries and two-thirds mints by weight and will sell for $1.25 per pound. How many pounds of each mixture should the candy manufacturer prepare in order to maximize his sales revenue?

Answer 1

Answer:

He should prepare 260 pounds of first mixture and 0 pounds of second mixture

Step-by-step explanation:

Let x be the total quantity ( in pounds ) of cherries and mints in the first mixture and y be the total quantity in second mixture,

Since, first mixture will contain half cherries and half mints by weight,

That is, in first mixture,

Cherries = $\frac{x}{2}$

Mints = $\frac{x}{2}$,

While, second mixture will contain one-third cherries and two-thirds mints by weight,

That is, in second mixture,

Cherries = $\frac{y}{3}$

Mints = $\frac{2y}{3}$

According to the question,

The manufacturer has 130 pounds of chocolate-covered cherries and 170 pounds of chocolate-covered mints in stock,

That is,

$\frac{x}{2}+\frac{y}{3} \leq 130$

$\frac{x}{2}+\frac{2y}{3}\leq 170$

Also, pounds can not be negative,

x ≥ 0, y ≥ 0,

Since, the first and second mixture must be sell at the rate of $2.00 per pound and $1.25 per pound respectively,

Hence, the total revenue,

Z = 2.00x + 1.25y

Which is the function that have to maximise,

By plotting the above inequalities,

Vertex of feasible regions are,

(0,255), (180, 120) and (260, 0),

Also, at (260, 0), Z is maximum,

Hence, he should prepare 260 pounds of first mixture and 0 pounds of second mixture in order to maximize his sales revenue.