Question 1112482: You are thinking of making your home 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 home has 80 light fittings and 1,500 square feet of uninsulated exterior wall. You can spend no more than $1,600 and would like to save as much per year in energy costs as possible. How many compact fluorescent light bulbs and how many square feet of insulation should you purchase?
compact fluorescent light bulbs =
insulation =
How much will you save in energy costs per year?
Answer by ikleyn(52855)   (Show Source):
You can put this solution on YOUR website! .
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You are thinking of making your home 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 home has 80 light fittings and 1,500 square feet of uninsulated exterior wall. You can spend no more than $1,600
and would like to save as much per year in energy costs as possible.
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?
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Let X = the number of compact fluorescent light bulbs and
let Y = "how many square feet of insulation should you purchase".
Then the "energy money" you will save per year is 2X + 0.2Y.
It is your "objective function" Z = 2X + 0.2Y.
The restrictions are:
4X + 1Y <= 1600 dollars (1) (<<<---=== money restriction)
0 <= X < = 80 (2) (<<<---=== the number of bulbs)
0 <= Y <= 1500 (3) (<<<---=== square feet)
Your feasibility area is shown in the Figure below:
Plot 4X + Y = 1600 (red), Y= 1500 (green) and X= 80 (blue)
Using the LP-method, you must check the objective function in the corner points of the feasibility area:
P1 = ( 0,1500)
P2 = (25,1500)
P3 = (80,1280)
P4 = (80,0)
You have
at P1 Z = 2*0 + 0.2*1500 = 300
at P2 Z = 2*25 + 0.2*1500 = 350
at P3 Z = 2*80 + 0.2*1280 = 416
at P4 Z = 2*80 + 0.2*0 = 160.
Thus the point P3 = (80,1280) = 80 bulbs and 1280 feet insulation is the solution.
Your expected annual energy saving will be 416 dollars.
By the way, your spending will be 4*80 + 1*1280 = 1600 dollars.
Answer. The optimal solution is 80 bulbs and 1280 square feet insulation.
Your expected energy saving will be 416 dollars.
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To see many other similar solved minimax problems by the LP-method, look into the lesson
- Solving minimax problems by the Linear Programming method
in this site.
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