Chapter 10.9, Problem 53E

Coffee Blends A coffee merchant sells two different coffee blends. The Standard blend uses 4 oz of arabica and 12 oz of robusta beans per package; the Deluxe blend uses 10 oz of arabica and 6 oz of robusta beans per package. The merchant has 80 lb of arabica and 90 lb of robusta beans available. Find a system of inequalities that describes the possible number of Standard and Deluxe packages the merchant can make. Graph the solution set.

To determine

The system of inequalities for the total number of standard and deluxe package the merchant make and sketch the graph for solution set.

Answer to Problem 53E

The system of inequalities that describes the production of standard and deluxe package by the merchant is $\{\begin{array}{l}\frac{1}{4}x+\frac{5}{8}y\le 80\\ \frac{3}{4}x+\frac{3}{8}y\le 90\\ x\ge 0,y\ge 0\end{array}$, and the graph is shown in Figure (1).

Explanation of Solution

Given:

Let the number of standard coffee package made by the merchant be x

Let the number of deluxe blend coffee made by the merchant be y

Total number of oz of arabica merchant contains is $1280$

Total number of robusta merchant contains is 1440

Total Arabica required by both standard and deluxe coffee should not greater than $1280$.

The first inequality is $4x+10y\le 1280$

Total robusta required by both standard and deluxe coffee should not exceed $1440$.

The second inequality is $12x+6y\le 1440$.

The number of standard and deluxe coffee is not negative

$x\ge 0$, $y\ge 0$

Take the first inequality $4x+10y\le 1280$

To plot the graph change the sign of inequality $4x+10y=1280$

Substitute 0 for x in $4x+10y=1280$

$y=128$

Substitute 0 for y in $4x+10y=1280$

$x=320$

Take the second inequality $12x+6y\le 1440$

To plot the graph change the sign of inequality $12x+6y=1440$

Substitute 0 for x in $12x+6y=1440$

$\begin{array}{c}6y=1440\\ y=240\end{array}$

Substitute 0 for y in $12x+6y=1440$

$\begin{array}{l}12x=1440\\ x=120\end{array}$

$x\ge 0,y\ge 0$ represents first quadrant.

The point of intersection between the first two inequality

Multiply the first equation $4x+10y=1280$ by 3.

$12x+30y=3840$

Subtract the second equation from first,

$\begin{array}{l}12x+30y=3840\\ \underset{\_}{12x+6y=1440}\\ 24y=2400\\ y=100\end{array}$

Substitute $100$ for y in $4x+10y=1280$

$\begin{array}{c}4x+1000=1280\\ 4x=1280-1000\\ x=\frac{280}{4}\\ x=70\end{array}$

Figure (1)

Thus, system of inequalities that describes the production of standard and deluxe package by the merchant are $\{\begin{array}{l}\frac{1}{4}x+\frac{5}{8}y\le 80\\ \frac{3}{4}x+\frac{3}{8}y\le 90\\ x\ge 0,y\ge 0\end{array}$, and the graph is shown in Figure (1).