Chapter 11, Problem 109RE

To determine

How many of each kind of engine is being produced in excess of the number that the factory is obligated to deliver?

Answer to Problem 109RE

  $\boxed{15}$ Gasoline engines

Explanation of Solution

Given information:

A factory produces gasoline engines and diesel engines. Each week the factory is obligated to deliver at least $20$ gasoline engines and at least $15$ diesel engines. Due to physical limitations, however, the factory cannot make more than $60$ gasoline engines nor more than $40$ diesel engines in any given week. Finally, to prevent layoffs, a total of at least $50$ engines must be produced. If gasoline engines cost $\$450$ each to produce and diesel engines cost $\$550$ each to produce, how many of each should be produced per week to minimize the cost? What is the excess capacity of the factory; that is, how many of each kind of engine is being produced in excess of the number that the factory is obligated to deliver?

Calculation:

The given linear programming problem requires that the production cost be minimized.

Let x represent the number of gasoline engines and y represents the number of diesel engines.

Since the factory is obligated to produce at least $20$ gasoline engines and at least $15$ diesel engines each week, the first two constraints are $x\ge 20$ and $y\ge 15$ .

The factory cannot produce more than $60$ gasoline and $40$ diesel engines due to physical limitations. So, the next two constraints are $x\le 60$ and $y\le 40$ . The total number of engines to be produced must be greater than or equal to $50$ . Thus, $x+y\ge 50$ .

Since the cost of each gasoline engine is $\$450$ , the cost $x$ gasoline engines is $450x$ . Similarly, $C=450x+550y$ , which is the objective function.

To minimize the given objective function, first graph the given inequalities and label the corner points.

  

List the corner points and the corresponding values of the objective function in a table.

  $\begin{array}{cc}\begin{array}{l}Corner-Po\mathrm{int}s\\ (x,y)\end{array}& \begin{array}{l}Values-of-the-objective-function\\ C=450x+550y\end{array}\\ (20,40)& C=450(20)+550(40)=31,000\\ (60,40)& C=450(60)+550(40)=49,000\\ (20,30)& C=450(20)+550(30)=25,500\\ (35,15)& C=450(35)+550(15)=24,000\\ (60,15)& C=450(60)+550(15)=35,250\end{array}$

From the table, we can see that the minimum value of the objective function is $24,000$ and it occurs at the point $(35,15)$ .

Hence, $35$ gasoline engines and $15$ diesel engines must be produced to minimize the cost.

The factory is obligated to produce gasoline engines not less than $20$ and diesel engines not less than $15$ , each week.

The amount of gasoline engines produced in excess id obtained by subtracting the required number from the total production of gasoline engines.

The number of gasoline engines produced is $35$ . Since only $20$ are needed, the factory is producing $35-20=15$ gasoline engines in excess.

The required number of diesel engines is $15$ and the factory’s production is also $15$ . So, there is no excess production for diesel engines.

Hence, the excess capacity of the factory is $15$ gasoline engines.