Q. 21

Question

Banquet Seating A banquet hall offers two types of tables for rent: 6-person rectangular tables at a cost of $$28$ each and 10 -person round tables at a cost of $$ 52$ each. Kathleen would like to rent the hall for a wedding banquet and needs tables for 250 people. The room can have a maximum of 35 tables, and the hall only has 15 rectangular tables available. How many of each type of table should be rented to minimize cost, and what is the minimum cost?

Step-by-Step Solution

Verified

Answer

The minimum cost is $$ 1252$ and it is achieved with 15 rectangular tables and 16 round tables.

1Step 1. Given information

Let us denote the number of 6 -person rectangular tables as x and the number of 10 -person round tables as y.
If C denotes the total cost of the tables to be rented, then C can be expressed as $C = 28 x + 10 y$ . This expression is the objective function.

2Step 2. Minimize the value of C subject to the two constraints x and y .

As there is only 15 rectangular tables available, we get the first constraint as $x \leq 15$ .
The hall can accommodate only 35 tables in total. Therefore, when x takes the value 15,y takes the value 20 . Therefore, the second constraint is $y \leq 20$ .

The total people to be seated are 250 in number in the two types of tables. Therefore, we get the third inequality as

6 x + 10 y \geq 250

.
The linear programming problem is to minimize

C = 28 x + 52 y

subject to the constraints ,

x \leq 15, y \leq 20

and

6 x + 10 y \geq 250

3Step 3. Graph the inequalities we obtained and label the corner points.

The shaded portion of the graph represents the set of feasible points.

Now, we have to find the value of the objective function at each corner point.

Corner point (x,y)	Value of cost $C = 28 x + 52 y$
$(8.5, 20)$	$C = 28 (8.5) + 52 (20) = 1278$
$(15, 20)$	$C = 28 (15) + 52 (20) = 1460$
$(15, 16)$	$C = 28 (15) + 52 (16) = 1252$