Question

FOUNDATIONAL ALGEBRA II

LINEAR PROGRAMMING 2
GET THE "CARRECT" ANSWER
NAME
The Gala Events Center has a rectangular parking lot measuring 20 m by 50 m. Only 60% of
the lot is usable space. A car requires 6 square meters of space and a bus requires 30 square
meters of space. The attendant can handle no more than 60 vehicles. If the parking fees are
$2.50 for cars and $7.50 for buses, how many of each type of vehicle should the attendant
accept to maximize income? What is the maximum income?
CLOSE TO "NUTTIN"
At "Nuttin' Like a Lightning Bolt Manufacturing" the cost to run Machine 1 for one hour is $2.00.
During that hour, Machine 1 produces 240 bolts and 100 nuts. The cost to run Machine 2 for an
hour is $2.40. During that hour, Machine 2 produces 160 bolts and 160 nuts. With a combined
running time of no more than 30 hours, how long should each machine run to produce an order
of at least 2080 bolts and 1520 nuts at the minimum operating cost?

Answer 1

1) The attendant should accept 15 cars and 45 buses to maximize the income, which would be $375.

2) Machine 1 should run for 8 hours and Machine 2 should run for 14 hours to produce an order of at least 2080 bolts and 1520 nuts at the minimum operating cost of $43.20.

The dimensions of the parking lot at the Gala Events Center = 20 m by 50 m

The total area of the parking lot = 1,000 m² (20 x 50)

Usable space = 60^% = 600 m²

Let the number of cars = x

Let the number of buses = y

The total number of vehicles parked = x + y

The maximum number of vehicles the attendant can handle is ≤ 60

Therefore, x + y ≤ 60

The area occupied by a car = 6 m²

The area occupied by a bus = 30 m²

The area occupied by x cars and y buses = 6x + 30y

The total income generated by parking x cars and y buses is 2.5x + 7.5y

To maximize the income, we need to maximize the expression 2.5x + 7.5y subject to the constraint x + y ≤ 60, using the Lagrange multipliers.

The Lagrangian function is:

L(x, y, λ) = 2.5x + 7.5y - λ(x + y - 60)

Taking partial derivatives concerning x, y, and λ:

∂L/∂x = 2.5 - λ = 0

∂L/∂y = 7.5 - λ = 0

∂L/∂λ = x + y - 60 = 0

Solving these equation:

x = 15

y = 45

Thus, the attendant should accept 15 cars and 45 buses to maximize the income.

The maximum income generated is 2.5(15) + 7.5(45) = $375.

2)

The number of bolts and nuts produced by Machine 1 in one hour = 240 bolts and 100 nuts

The number of bolts and nuts produced by Machine 2 in one hour = 160 bolts and 160 nuts

The cost to run Machine 1 for one hour = $2.00

The cost to run Machine 2 for one hour = $2.401

Let the number of hours that Machine 1 runs = x

Let the number of hours that Machine 2 runs = y

To minimize the operating cost, we use the following equation:

C = 2x + 2.4y

The constraints are:

240x + 160y ≥ 2080 (minimum number of bolts)

100x + 160y ≥ 1520 (minimum number of nuts)

x + y ≤ 30 (maximum running time)

Using linear programming, the solution is:

x = 8

y = 14

Thus, Machine 1 should run for 8 hours and Machine 2 should run for 14 hours to produce an order of at least 2080 bolts and 1520 nuts at the minimum operating cost.

The minimum operating cost is:

C = 2(8) + 2.4(14) = $43.20.