# Maximum Area Application Wanda and Louise Maximum Area

• Slides: 14

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 = 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 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 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? 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 50 0 0 5 10 15 20 25 30 35

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!!