Maximum Area Application Wanda and Louise Maximum Area

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Maximum Area - Application Wanda and Louise Maximum Area

Maximum Area - Application Wanda and Louise Maximum Area

The Problem Wanda and Louise raise puppies. They need a rectangular fenced enclosure for

The Problem Wanda and Louise raise puppies. They need a rectangular fenced enclosure for the puppies to run and play. A contractor said it would cost $30/m of fencing and they have $480 to spend. $480 ÷ $30 / m = 16 m They can afford 16 m of fencing! What is the maximum area they can enclose?

Why is this Quadratic? Perimeter (m) 16 Length (m) 1 2 3 4 5

Why is this Quadratic? Perimeter (m) 16 Length (m) 1 2 3 4 5 6 7 Width (m) 7 6 5 4 3 2 1 Graph Length vs Area! Area (m 2) 7 12 15 16 15 12 7

Graph of Length(m) vs Area (m 2) Width (m) vs Area (m 2) would

Graph of Length(m) vs Area (m 2) Width (m) vs Area (m 2) would be identical!!

Now let’s solve the problem!

Now let’s solve the problem!

Part II – Against a House Wanda and Louise Maximum Area

Part II – Against a House Wanda and Louise Maximum Area

Part III – Two Separate Enclosures Wanda and Louise Maximum Area

Part III – Two Separate Enclosures Wanda and Louise Maximum Area

Revenue – Application Problems T-Shirt Revenue Page 273

Revenue – Application Problems T-Shirt Revenue Page 273

The Problem Mirna operates her own store, Mirna’s Fashion. A popular style of T-shirt

The Problem Mirna operates her own store, Mirna’s Fashion. A popular style of T-shirt sells for $10. At that price, Mirna sells about 30 T-shirts a week. Experience has taught Mirna that changing the price of an article has an effect on sales. For example, she knows that a $1 increase in the price of the T-shirt means that she will sell about one less T-shirt per week. Mirna wants to find the price that will maximize her revenue from the sale of T-shirts!!

Why is this Quadratic? Price ($) Number Sold 8 32 Revenue ($) (price)(# sold)

Why is this Quadratic? Price ($) Number Sold 8 32 Revenue ($) (price)(# sold) 256 9 31 279 10 30 300 11 29 319 12 28 336 13 27 351 14 26 364 Graph Price vs Revenue. . .

Price ($) vs Revenue ($) Revenue 450 400 350 300 250 200 150 100

Price ($) vs Revenue ($) Revenue 450 400 350 300 250 200 150 100 50 0 0 5 10 15 20 25 30 35

Now let’s solve the problem!

Now let’s solve the problem!

Another Example A ticket to the school dance costs $6. At this price, 250

Another Example A ticket to the school dance costs $6. At this price, 250 students attended. The dance committee knows that for every $1 increase in price, 25 fewer tickets are sold. • What ticket price maximizes the revenue? • What is the amount of revenue taken in?

Homework Solve each example by expanding and completing the square to get vertex form

Homework Solve each example by expanding and completing the square to get vertex form instead!!