Quiz 10 (Solution)

Date: MATH 1090-4 - Fall 2003

1. Read the following problem, but do not solve it!

   A privately owned lake contains two types of game fish, bass and trout. The owner provides two types of food, A and B, for these fish. Bass require $2$ units of food A and $4$ units of food B, and trout require $5$ units of food A and $2$ units of food B. If the owner has $800$ units of each food, find the maximum number of fish that the lake can support.

   1. What variables would you use to describe the problem?

      From the problem we see that we want to describe the number of fish, and, since there are $2$ types of fish we use two variables, one to count each type.

      $b=$ number of bass in the lake.

      $t=$ number of trout in the lake.

   2. Write the feasible region to be used to solve the problem (that is, write down the inequalities but do not graph!).

      Clearly, both $b$ and $t$ must be $\geq 0$ since they represent numbers of fish. For food A, since bass require $2$ units and trouts $5$ units and, all together there are at most $800$ units we must have $2b+5t\leq 800$. Similarly, for food B we have $4b+2t\leq 800$. So, the feasible region determined by the problem is:

      $$\begin{cases}b\geq 0\ t\geq 0\ 2b+5t\leq 800\ 4b+2t\leq 800. \end{cases}$$

      
      

   3. Write the function that you want to maximize or minimize in this problem.

      The problem asks to maximize the number of fish. Therefore, we are interested in the total number of fish ($b$+$t$), and the function is

      $$N(b,t) = b+t.$$

      
      

   4. Do you want to maximize or minimize the function that you wrote in the previous item?

      We want to maximize the amount of fish, that is, $N(b,t)$.

2. Consider $f(x,y)=2x-y$ and the feasible region

   $$\begin{cases}x\leq 10\ y\geq 0\ 2x-3y\geq -4\ 2x+y\geq 12. \end{cases}$$

   
   
   1. Sketch the graph of the feasible region showing clearly the coordinates of each corner.

      We consider each inequality separately first. The border of the region determined by the first one is the vertical line $x=10$ and the corresponding half-plane sits to the left of that line. For the second inequality, the border is the line $y=0$ (this is the $x$-axis) and the half-plane is above that line.

      The third inequality leads to the border line $2x-3y=-4$ that contains, for instance, the points $(-2,0)$ and $(0,12)$; by testing the point $(0,0)$ we see that the half-plane lies below the line. Finally, the border of $2x+y\geq 12$ is the line $2x+y= 12$, that contains the points $(6,0)$ and $(0,12)$; by testing the point $(0,0)$ we conclude that the half-plane is to the right of that line.

      The feasible region is shown in Figure 1. All four border lines are contained in the corresponding half-plane, so that they appear solid in the figure.

      Figure 1: Feasible region for exercise 2

      

      The corners are found by finding the intersection of the corresponding lines. The two corners on the $x$-axis have $y=0$ and, plugging this into the equations for the line $I$ and $IV$ ($x=10$ and $2x+y= 12$) we obtain the values $x=10$ and $x=6$, and the corners are $(10,0)$ and $(6,0)$ respectively. The intersection of the lines $I$ and $III$ is found by plugging $x=10$ into $2x-3y=-4$, that gives $y=8$, so that the corner is $(10,8)$. Finally, the intersection of $III$ and $IV$ is the solution of the system $\begin{cases} 2x-3y=-4\ 2x+y=12 \end{cases}$ that gives $x=4$, $y=4$, so that the corresponding corner is $(4,4)$.

   2. Find the maximum and minimum values of $f(x,y)$ over the feasible region. If any of these values does not exist, explain why.

      We see that the feasible region is closed and bounded, so it suffices to evaluate $f$ at the four corners and the highest value will be the maximum, while the lowest is the minimum.

      Since $f(6,0)=12$, $f(4,4)=4$, $f(10,8)=12$, and $f(10,0)=20$, we see that the maximum value of $f$ is $20$ (realized at $(10,0)$) and the minimum value is $4$ (realized at $(4,4)$).