Solutions to Quadratic Equations Word Problems

Transcription

1 Area Problems: Solutions to Quadratic Equations Word Problems 9. A local building code requires that all factories must be surrounded by a lawn. The width of the lawn must be uniform and the area must equal the area of the factory. What must be the width of the lawn surrounding a rectangular factory that measures 120 m by 80 m? Area factory = 120m 80m Area lawn = Area factory Therefore, Area lawn = Area factory = 120m 80 m = 9600 m 2 Unknown information: width of the lawn Let the width of the lawn be Area lawn = Area total - Area lawn 9600 = ( )( ) = ( ) = = 0 4( ) = 0 4( +120)( -20) = Therefore ( + 120) = 0 or ( 20) = 0 Therefore = -120 or = 20 Since is the width, -120 is inadmissible, therefore = 20 Therefore, the width of the lawn is 20 m 10. A factory is to be built on a lot that measures 80m 60m. A lawn of uniform width and equal in area to the factory must surround the factory. What dimensions must the factory have? Area lot = Area lawn + Area factory = 80m 60m = 4800 m 2 Area lawn = Area factory Unknown information: dimensions of the factory 60 Let the width of the lawn be Area factory = Area lot - Area lawn (80 2)(60-2) = (80 2)(60-2) 2(80 2)(60-2) = ( ) = = = = 0 8( ) = 0 8( - 60) ( 10) = 0 Therefore ( - 60) = 0 or ( 10) = 0 Therefore = 60 or = 10 Since represents the width, = 60 is inadmissible, therefore = The dimensions of the factory are (80-2(10)) (60-2(10)) = Therefore, the dimensions of the factory are 60m 40m

2 12. A matte of uniform width is to be placed around a painting so that the area of the matted surface is twice the area of the picture. If the outside dimensions of the matte are 40 cm and 60 cm, find the width of the matte. Area matte = 2 Area picture Area total = 40 cm 60 cm = 2400 cm 2 Unknown information: width of the matte Let the width of the matte be Area matte = Area total - Area picture 2(Area picture ) = Area total - Area picture 40 3(Area picture ) = Area total 3(60 2) (40 2) = ( ) = = = 0 12( ) = 0 24( 40)( 10) = 0 60 Therefore ( - 40) = 0 or ( 10) = 0 Therefore = 40 or = 10 Since is the width, = 40 is inadmissible, therefore = 10 Therefore, the width of the matte is 10 cm 14. A picture 20cm wide by 10cm high is to be centrally mounted on a rectangular frame with total area three times the area of the picture. Assuming equal margins for all four sides, find the width of the margin. Area total = 3 Area picture Area picture = 20 cm 10 cm = 200 cm 2 Unknown information: width of the margin Let the width of the margin be (Area picture ) = Area total 3(20)(10) = (20 + 2)(10 + 2) 600 = ( ) 600 = = 0 4( ) = 0 4( + 20)( 5) = 0 Therefore ( + 20) = 0 or ( 5) = 0 Therefore = -20 or = 5 Since is the width, = -20 is inadmissible, therefore = 5 Therefore, the width of the margin is 5 cm

3 Merchandising Problems: 21. The Puck Heads hockey team averages people at each hockey game with the average price of a ticket being $60. In the championship game, admission prices will increase. For every $5 increase in ticket price, 1000 fewer people will attend. What price will maimize the Puck Heads revenue? Let be the number of times the price is increased by $5. Price per ticket = (60 + 5) Tickets Sold = ( ) Revenue = Price Tickets Sold Revenue = (60 + 5) ( ) To find the maimum revenue, you must find the verte. To do this you must first find the zeros. 0 = (60 + 5) ( ) Therefore (60 + 5) = 0 and ( ) = 0 Therefore = -12 and = 15 The co-ordinate for maimum revenue is = ( )/2 = 3/2 The new price is ( (3/2)) = Therefore, the price to maimize the revenue is $ The local Transit commission s single fare price is $1.20 cash. On a typical day, 240,000 people take the transit and pay the single-fare price. To reflect higher costs, single fare prices will be increased but surveys show that every $0.05 increase in fare will reduce ridership by 8000 riders daily. What single-fare price will maimize revenue for the commission based on single fares? Let be the number of times the price is increased by $0.05. Price per ticket = ( ) Tickets Sold = ( ) Revenue = Price Tickets Sold Revenue = ( ) ( ) To find the maimum revenue, you must find the verte. To do this you must first find the zeros. 0 = ( ) ( ) Therefore ( ) = 0 and ( ) = 0 Therefore = -24 and = 30 The co-ordinate for maimum revenue is = ( )/2 = 3 The new price is ( (3)) = 1.35 Therefore, the price to maimize the revenue is $1.35

4 23. A company sells 9000 pairs of slacks a month at an average of $70 each. The store is going to increase prices in order to increase profits. Sales forecasts indicated that sales will drop by 200 for every dollar increase in price. On average, the company pays $30 for each pair of slacks that it sells. What price will maimize the profits? Let be the number of times the price is increased by $1. Price per pair of slacks = (70 + ) Pants Sold = ( ) Revenue = Price Slacks Sold Revenue = (70 + ) ( ) Cost = 30( ) To find the maimum profit: Profit = Revenue Cost P = (70 + )( ) 30( ) = ( )( ) = ( )(40 + ) Set P = 0 Therefore 0 = ( )(40 + ) = 0 or 40 + = 0 X = 45 = -40 The co-ordinate for maimum profit is: New Price = = $ Therefore, the price to maimize the profits is $ A hardware store currently sells light bulbs each week at a price of $7.00 per package. To increase sales and reach more customers, the store decides to reduce the price of the package knowing that every $0.10 decrease in price will result in 5 more sales. What price will maimize total revenue? Let be the number of times the price is decreased by $0.10. Price per light bulb package = (7 0.1 ) Light Bulbs Sold = ( ) Revenue = Price Light Bulbs Sold Revenue = (7 0.1) ( ) To find the maimum revenue, you must find the verte. To do this you must first find the zeros. 0 = (7 0.1) ( ) Therefore (7-0.1) = 0 and ( ) = 0

5 Therefore = 70 and = - 60 The co-ordinate maimum revenue is = (70-60)/2 = 5 The new price is (7 0.1 (5)) = 6.50 Therefore, the price to maimize the revenue is $6.50