Download Solving problems with Quadratic Relations

MPM2D
Unit 3 – Quadratic Applications
Solving problems with Quadratic Relations. Sample types of problems
Type I - Projectile Motion problems where you are given the equation.
Example. p316 #9. A stunt diver follows a path of H= -4.9t2 + 21. Find the time when he is
halfway down. Find the time when he hits the water.
Solution: a) To find halfway point set H =10.5 and solve
10.5 =-4.9t 2 + 21
4.9t 2 = 10.5
10.5
4.9
t  1.46
t 
Time must be positive, so the diver is halfway down in about 1.5 seconds.
b) To find when he hits the water set H = 0 and solve 0 =-4.9t2 + 21
0 =-4.9t 2 + 21
4.9t 2 = 21
21
4.9
t  2.07
t 
Time must be positive, so the diver hits the water in about 2.1 seconds.
The time is less than double the previous time due to acceleration.
Type II - Projectile Motion problems where you are given the zeros
Jon fires a toy rocket into the air from the ground. The rocket is in the air for 10 s before it falls to
the ground. Jon knows that after 2 s, the rocket was 20m in the air. Find the greatest height of the
rocket.
Let y be the height of the rocket.
Let x be the time.
The zeros are 0 and 10. so in factored form the equation will be
y = a(x – 0)(x – 10)
Substitute in known point (2, 20)
20 = a(2 – 0)(2 – 10)
20 = -16a
a = -1.25
[note it makes sense that a is negative, since parabolas opens down]
Equation is y = - 1.25(x)(x – 10)
The great height will be at 5 s which is the midpoint of the two zeros 0 and 10.
Substitute in x = 5
y = -1.25(5)(5- 10)
y = 31.25
Therefore the greatest height is 31.25 m
MPM2D
Unit 3 – Quadratic Applications
Type III - Revenue Problems
A restaurant determines that each 10 cent increase in the price of a salad results in 25 fewer
salads being sold. The usual price for a salad is $2.00 and the restaurant sells 300 salads each
day.
a) Write an equation to model the daily revenue for the restaurant.
b) Find the optimum price for a salad
c) What is the optimal revenue?
Solution:
Assume that when the price decreases by 10 cents they will sell 25 more salads.
Let x be the number of price increases of 10 cents
Let y be the revenue
a) Revenue = Price x Quantity
y = (2.00 + 0.10x)(300 – 25x)
b) The zeros will be when 2.00 + 0.10x = 0 and 300 – 25x = 0
x = -20 and x = 12
The midpoint is x = -4
There the optimal price will be 4 price decreases. 2.00 + 0.10(-4) = 1.60
The optimal price will be $1.60
c) The axis of symmetry is x = -4 from above. Plug in to find optimal value
y = (2.00 + 0.10x)(300 – 25x)
y = (2.00 + 0.10(-4))(300 – 25(-4))
y = 640
The optimal revenue will be $640, when the price is $1.60 they sell 400 salads.
Type IV - Border Problems
A garden planted in the middle of a 40m by 30m field. A border of flowers are planted around the
garden so that the area of the flowers is exactly one half the area of the field. The width of the
border is constant. Find the dimensions of the garden.
Solution:
The area of the field is 1200m 2
The x represent the width of the border.
The equation area of the garden will be
Area = (30 – 2x)(40 – 2x)
Area needs to be ½ the area of the field so will be
600m2
600 = (30 – 2x)(40 – 2x)
600 = 1200 – 140x – 4x2
4x2+140x – 600 = 0 (divide by 4)
x2 + 35x – 150 = 0 (factor)
(x + 30)(x + 5) = 0
so x = -30 or x= 5
since distance is positive x = 5. The border is 5m.
Thus the dimensions of the garden is 30m by 20m