APPLICATION PROBLEMS OF QUADRATIC FUNCTION

Problem 1 :

A stone is thrown into the air and its height in meters above the ground is given by the function

h(t) = -5t² + 30t + 2

where t is the time (in seconds) from when the stone is thrown.

a) How high above the ground is the stone at time t = 3 seconds ?

b)  How high above the ground was the stone released ?

c)  At what time was the stone's height above the ground 27 m ?

Solution :

h(t) = -5t² + 30t + 2

a) Applying t = 3, we get

h(3) = -5(3)² + 30(3) + 2

h(3) = -5(9) + 90 + 2

= -45 + 92

= 47 m

So, height of the stone is 47 m above the ground.

b) When the stone is released t = 0

h(t) = -5t² + 30t + 2

h(0) = -5(0)² + 30(0) + 2

h(0) = 2 m

When the stone is released 2 m above ground level.

c)  When h(t) = 27, t = ?

27 = -5t² + 30t + 2

-5t² + 30t + 2 - 27 = 0

-5t² + 30t  - 25 = 0

Dividing by -5, we get

t² - 6t  + 5 = 0

(t - 1)(t - 5) = 0

Equating each factor to 0, we get

t = 1 and t = 5

After 1 sec and 5 sec, the height is 27 m.

Problem 2 :

An object is projected into the air with a velocity of 30 m/s. Its height in meters, after t seconds is given by the function 

h(t) = 30t - 5t²

a) Calculate the height after 

i) 1 second     ii)  5 seconds     iii)  3 seconds

b)  Calculate the times at which the height is 

i) 40 m   ii)  0 m

c) Explain your answer is part b.

Solution :

h(t) = 30t - 5t²

a) 

iii) When t = 3

h(3) = 30(3) - 5(3)²

= 90 - 5(9)

= 90 - 45

= 45 m

b)  Calculate the times at which the height is 

i)  When h = 40 m

40 = 30t - 5t²

5t² - 30t + 40 = 0

Dividing by 5, we get

t² - 6t + 8 = 0

(t - 2)(t - 4) = 0

t = 2 and t = 4

ii)  When h = 0 m

0 = 30t - 5t²

5t² - 30t = 0

Dividing by 5, we get

5t(t - 6) = 0

t = 0 and t = 6

c) After 2 second or 4 seconds, the height of the object will be 40 m

After 0 seconds or 6 seconds, the height of the object is 0 m or will be on the land.

Problem 3 :

A cake manufacturer finds that the profit in dollars of making x cakes per day is given by the function

P(x) = (-1/4)x² + 16x - 30

a) Calculate the profit if 

i) 0 cakes     ii)  10 cakes are made per day 

b) How many cakes per day are made if the profit is $57 ?

Solution :

P(x) = (-1/4)x² + 16x - 30

a) Number of cakes - x

i) When x = 0

P(0) = (-1/4)(0)² + 16(0) - 30

= -30

When 0 cakes lost, loss happened of $30.

ii)  When x = 10

P(10) = (-1/4)(10)² + 16(10) - 30

= -25 + 160 - 30

= -55 + 160

= 135

When 0 cakes lost, loss happened of $30.

b) When profit = $57

57 = (-1/4)x² + 16x - 30

228 = -x² + 64x - 120

-x² + 64x - 120 - 228 = 0

-x² + 64x - 348 = 0

x² - 64x + 348 = 0

(x - 58)(x - 6) = 0

x = 58 and x= 6

When profit made of $52, the number of cakes is 58 or 6.

Problem 4 :

The height H meters, of a cannon ball t seconds after it is fired into the air is given by

H(t) = -4t² + 16t + 9

a)  Find the time taken for the cannon ball to reach the maximum height.

b) what is the maximum height reached by the cannon ball.

c)  How long does it take for a cannonball to fall back to earth ?

Solution :

H(t) = -4t² + 16t + 9

a)  To find the maximum height, we have to find the vertex of the parabola

x-coordinate of the vertex = -b/2a

a = -4, b = 16 and c = 9

x = -16/2(-4)

x = 16/8

x = 2

To reach the maximum height it will take 2 seconds.

b)  Applying x = 2, we get

H(t) = -4(2)² + 16(2) + 9

= -4(4) + 32 + 9

= -16 + 41

= 25

So, the maximum height is 25 m.

c) When the ball reaches the ground, the height will be 0.

0 = -4t² + 16t + 9

-4t² + 16t + 9 = 0

4t² - 16t - 9 = 0

4t² - 18t + 2t - 9 = 0

2t(t - 9) + 2(t - 9) = 0

(2t + 2)(t - 9) = 0

t = -1 and t = 9

So, after 9 seconds the ball will reach the ground.