Quadratic Applications

1. Quadratic Applications

2. Mr. Jones has 56 feet of fencing to make a rectangular dog pen. If he just uses his house to be one side of the pen, what would be the length and width for the maximum area of the dog pen? House Area = l*w Our equation would be: y= x(56-2x) Find the x coord. of the vertex to get the width. Then plug it in to get length! Area x x (50-2x)

3. A store sells 32 pairs of jeans a day for $40 each. The owner thinks for each $2 increase in price, the number of jeans sold will go down by 1. What price should be charged to maximze profit? - my goal here is to write an equation, then find the vertex. (# jeans sold)(price) “ x is for each increase in price” (32-1x)(40+2x) Prof I t # of increases The x-coord of the vertex is the average of your solutions.. So solve each parenthesis and average that #. “number of jeans goes down by 1” “$2 increase in price” The x-coord of the vertex will be the “maximum” number of price increases. Once you get that value, plug in into your price parenthesis!

4. (# jeans)(price) (32-1x)(40+2x) 32-1x=0 X=32 40+2x=0 X=-20 Plug this into the “price” parenthesis.. $52 is the price that would maximize the profit

5. If an object is propelled upward from a height of 96ft at an initial velocity of 80ft per second, then its height after t seconds is given by the equation: where h is in feet. a) What is the maximum height? b) At what time will it reach that height? c) When will the object hit the ground? He i g h t To get a) and b) find the vertex!! Maximum height = y – coord. What time it reaches the max height = x-coord time To see when the object hits the ground, set the equation equal to zero and solve!! To find the vertex from standard form… Then plug in to get y Use the quad formula. You’ll get two answers. Just use the positive one! Vertex (2.5,196) t=6