Maximum Area - Application

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Presentation transcript:

Maximum Area - Application Wanda and Louise Maximum Area

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 = 16m They can afford 16m of fencing! What is the maximum area they can enclose?

Why is this Quadratic? Graph Length vs Area! Perimeter (m) Length (m) Width (m) Area (m2) 16 1 7 2 6 12 3 5 15 4 Graph Length vs Area!

Graph of Length(m) vs Area (m2) Width (m) vs Area (m2) would be identical!!

Now let’s solve the problem!

Part II – Against a House Wanda and Louise Maximum Area

Part III – Two Separate Enclosures Wanda and Louise Maximum Area

Revenue – Application Problems T-Shirt Revenue Page 273

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

Price ($) vs Revenue ($)

Now let’s solve the problem!

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 instead!!