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You are thinking of making your mansion more energy efficient by replacing some of the light bulbs with compact fluorescent bulbs, and insulating part or all of your exterior walls. Each compact fluorescent light bulb costs $4 and saves you an average of $2 per year in energy costs, and each square foot of wall insulation costs $1 and saves you an average of $0.20 per year in energy costs.? Your mansion has 300 light fittings and 2800 sqft of uninsulated exterior wall. To impress your friends, you would like to spend as much as possible, but save no more than $800 per year in energy costs (you are proud of your large utility bills). How many compact fluorescent light bulbs and how many square feet of insulation should you purchase? How much will you save in energy costs per year?

Sagot :

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, [tex]C_s[/tex] = 2·x + 0.2·y

Where;

x = 120, and y = 2,800

∴ [tex]C_s[/tex] = 2 × 120 + 0.2 × 2,800 = 800

The amount of savings in energy cost per year, [tex]C_s[/tex] = $800

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