Chapter 6.5, Problem 31PPE

To determine

To find: To set and graph a linear inequality on modeling given situation and then finding three possible combinations of amounts of each fish to be sold.

Answer to Problem 31PPE

Required linear inequality is

  $9x+12y\ge 120$ and its graph is,

  

Three possible combinations of both type of fishes are,

1. 10 pounds of cod and 20 pounds of flounder.
2. 20 pounds of cod and 30 pounds of flounder.
3. 30 pounds of cod and 20 pounds of flounder.

Explanation of Solution

Given information: Cod fish costs $9 per pound and flounder costs $12 pound. Both type of fishes to be bought in a budget of at-least $120.

Concept used: First set the linear inequality, representing given situation using following statement,

  $\text{Amount of cod fish }\times \text{its cost+amount of flounder fish}\times \text{its cost}\ge \text{120}$

And then find three values of unknown variables, satisfying this inequality.

Calculation: Let x pounds of cod and y pounds of flounder fishes are at-least bought. So, use above statement to form linear inequality as,

  $9x+12y\ge 120$

Calculate its two points of related equation $9x+12y=120$ in ordered pair form as follows,

x	$9x+12y\ge 120$	$(x,y)$
0	$\begin{array}{l}9\times 0+12y=120\\ 12y=120\\ y=10\end{array}$	$(0,10)$
8	$\begin{array}{l}9\times 8+12y=120\\ 12y=120-72=48\\ y=\frac{48}{12}=4\end{array}$	$(8,4)$

Plot these points in XY plane and get the given inequality satisfied by point (0,0), so that to find its solution region as shown in below graph,

  

Three possible amounts of fishes, as shown in above shaded area, are,

  $\begin{array}{l}(i)x=10,y=20\\ (ii)x=20,y=30\\ (iii)x=30,y=20\end{array}$

Conclusion: So, different possibility of buying may be,

1. 10 pounds of cod and 20 pounds of flounder.
2. 20 pounds of cod and 30 pounds of flounder.
3. 30 pounds of cod and 20 pounds of flounder.