Linear System Applications: Break Even Point and Water Tanks Filling
Explore applications of linear systems in determining the break-even point for a fashion designer selling hats and finding when two water tanks will have the same amount of water. Understand concepts through equations and graphical solutions. Complete assigned problems for practice.
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Section 6.4 Applications of Linear System
A fashion designer makes and sells hats. The material for each hat costs $5.50. The hats sell for $12.50 each. The designer spends $1400 on advertising. How many hats must the designer sell to break even? Define Variables: x = number of hats sold y = the # of dollars of expense or income Write a system of equations. Income: Expense: y = 12.50x or y = 12.5x y = 5.5x + 1400 Choose a method: graph, substitution, elimination. y = 5.5x + 1400 ( ) = 5.5x + 1400 12.5x 12.5x = 5.5x + 1400 7x = 1400 Answer? x = 200 200 hats
The local zoo is filling two water tanks for the elephant exhibit. One water tank contains 50 gal of water and is filled at a constant rate of 10 gal/h. The second water tank contains 29 gal of water and is filled at a constant rate of 3 gal/h. When will the two tanks have the same The local zoo is filling two water tanks for the elephant exhibit. One water tank contains 50 gal of water and is filled at a constant rate of 10 gal/h. The second water tank contains 29 gal of water and is filled at a constant rate of 3 gal/h. When will the two tanks have the same amount of water? Explain. amount of water? Explain. Let h = the number of hours the tanks are filling. Write a system of equations. Let g = the number of gallons in the tank. Tank 1: Tank 2: g = 10h + 50 g = 3h + 29 Solve the system solve for g: g = 10h + 50 Hours = -3 Gallons = 20 ( ) = 10h + 50 3h + 29 g = 10h + 50 3h + 29 = 10h + 50 Answer? g = 10( ) + 50 -3 -7h+ 29 = 50 -7h = 21 g = -30 + 50 g = 20 Never: it is impossible to have time be -3 hours. h = -3
Assignment: Pg 390: 7-10, 13-16, 19-21
