A furniture store has set aside 800 square feet to display its sofas and chairs. Each sofa utilizes 50 sq. ft. and each chair utilizes 30 sq. ft. At least five sofas and at least five chairs are to be displayed. a. Write a mathematical model representing the store's constraints.
b. Suppose the profit on sofas is $200 and on chairs is $100. On a given day, the probability that a displayed sofa will be sold is 0.03 and that a displayed chair will be sold is 0.05. Mathematically model each of the following objectives:

1. Maximize the total pieces of furniture displayed.
2. Maximize the total expected number of daily sales.
3. Maximize the total expected daily profit.

a) 50S + 30C ≤ 800

b) 1) MAX = S + C

   2) Max = 0.03S + 0.05C

   3) Max = 6S + 5C

Step-by-step explanation:

Given:

Total space = 800 square feet

Each sofa = 50 square feet

Each chair = 30 square feet

At least 5 sofas and 5 chairs are to be displayed.

a) Write a mathematical model representing the store's constraints:

Let S denote number of sofas displayed and C denote number of chairs displayed.

The mathematical model will be:

50S + 30C ≤ 800

At least 5 sofas are to be dispayed: S ≥ 5

At least 5 chairs are to be displayed: C ≥ 5

b)

1) Maximize the total pieces of furniture displayed:

S + C = MAX

2) Maximize the total expected number of daily sales:

MAX = 0.03S + 0.05C

3) Maximize the total expected daily profit:

Given:

Profit on sofas = $200

Profit on chairs = $100

Max Expected daily profit =

Max = (200S * 0.03) + (100C * 0.05)

Max = 6S + 5C

- "runner b is moving at a slower rate."

- "the graphs shown are graphs of distance(km) against time(min).

the slopes of this graphs are are given by:

slope = (change in y)/(change in x)

= (change in distance)/(change in time)

this is the formula for finding the speed. so, the slope of these graphs represent the speed of the runner.

taking each division in the graph to represent 1 on both axis:

runner a ⇒ speed = (2-0)/(1-0)

= 2/1

= 2 km/min

runner b ⇒ speed = (3-0)/(3-0)

= 3/3

= 1 km/min

this shows that   runner b is moving at a slower rate."