Answer:
1) The number of compact fluorescent light bulbs to buy = 120 bulbs
The number of square feet of insulation to be purchased = 2,800 sqft
2) The amount of savings in energy cost per year = $800
Step-by-step explanation:
The cost of each compact fluorescent bulb = $4
The average amount saved by each bulb per year = $ 2
The cost of each square foot of wall insulation = $ 1
The amount saved by each square foot of wall insulation = $ 0.20
The number of light fittings in the mansion = 300 fittings
The area of uninsulated exterior wall of the mansion = 2,800 ft.²
The maximum amount saved in energy cost = $800
1) Let 'x' represent the number of compact fluorescent bulb used and let 'y' represent the number of square feet of insulation purchased
We have;
2·x + 0.2·y ≤ 800
∴ y ≤ 4,000 - 10·x
Total Cost, P = 4·x + y
From the combined graph of the total cost and cost savings, we have the highest total cost is given when the wall insulation area, y is maximum, that is y = 2,800 and x = 120
Therefore, we have;
Total Cost, P = 4·x + y
∴ P = 4 × 120 + 2,800 = 3,280
Therefore, the number of compact fluorescent light bulbs and the number of square feet of insulation to be purchased are 120 bulbs and 2,800 square feet of insulation respectively
2) The amount of savings in energy cost per year, = 2·x + 0.2·y
Where;
x = 120, and y = 2,800
∴ = 2 × 120 + 0.2 × 2,800 = 800
The amount of savings in energy cost per year, = $800