QUADRATIC MAXIMUM AND MINIMUM WORD PROBLEMS

Problem 1 :

The function y = -16t² + 248 models the height y in feet of a stone t seconds after it is dropped from the edge of a vertical cliff. How long will it take the stone to hit the ground ? Round to the nearest hundredth of a second.

a. 7.87 seconds       c. 3.94 seconds

b. 0.25 seconds        d. 5.57 seconds

Solution :

Given, y = -16t² + 248

When y = 0

0 = -16t² + 248

248 = 16t²

248/16 = t²

√248/16 = t

√31/2 = t

√15.5 = t

3.94 = t

So, 3.94 seconds long will it take the stone to hit the ground.

So, option (c) is correct.

Problem 2 :

y = 3x² – 36x + 111

What is the minimum value of the function ?

Solution :

To get maximum or minimum value of the quadratic function, we have to write it in the vertex form.

y = 3(x² – 12x) + 111

y = 3(x² – 2 ⋅ 6 ⋅ x + 6² - 6²) + 111

y = 3[(x – 6)² - 6²] + 111

y = 3[(x – 6)² - 36] + 111

y = 3(x – 6)² - 108 + 111

y = 3(x – 6)² + 3

a = 3 > 0, the parabola opens up. So, it will have minimum value.

Vertex is at (6, 3)

So, the minimum value is at y = 3.

Problem 3 :

A manufacturer determines that the number of drills it can sell is given by the formula D = -4p² + 160p – 305, where p is the price of the drills in dollars.

a. At what price will the manufacturer sell the maximum number of drills?

b. What is the maximum number of drills that can be sold ?

a. $80; 13,105 drills  c. $40; 305 drills 

b. $22; 1,289 drills    d. $20; 1,295 drills 

Solution :

a. D = -4p² + 160p – 305

P = -b/2a

= -160/2(-4)

= -160/-8

p = 20

So, $20 price will the manufacturer sell.

b. Maximum number of drills:

 D = -4p² + 160p – 305

= -4(20)² + 160(20) – 305

= -1600 + 3200 – 305

= 1295

So, the maximum number of drills is 1295.

So, option (d) is correct.

Problem 4 :

After t seconds, a ball tossed in the air from the ground level reaches a height of h feet given by the function

h(t) = 144t – 16t²

a. What is the height of the ball after 3 seconds?

b. What is the maximum height the ball will reach?

c. After how many seconds will the ball hit the ground before rebound?

Solution :

h(t) = 144t – 16t²

Here h(t) is height, and t is time taken.

a) h(t) = ?, when t = 3

h(3) = 144(3) – 16(3)²

= 432 - 144

= 288

b) To find maximum height, we have to write the quadratic equation in vertex form.

h(t) = -16t² + 144t

h(t) = -16t² + 144

h(t) = -16(t² - 9t)

h(t) = -16[t² - 2 ⋅ t ⋅ 9/2 + (9/2)² - (9/2)²]

h(t) = -16[(t - (9/2))² - (81/4)]

h(t) = (t - (9/2))² + 324

Maximum height is 324 ft.

c) When height = 0

-16t² + 144t = 0

-16t(t - 9) = 0

t = 0 and t = 9

Problem 5 :

A rocket carrying fireworks is launched from a hill 80 feet above a lake. The rocket will fall into the lake after exploding at its maximum height. The rocket’s height above the surface of the lake is given by the function

h(t) = -16t² + 64t + 80.

a. What is the height of the rocket after 1.5 seconds?

b. What is the maximum height reached by the rocket?

c. After how many seconds after it is launched will the rocket hit the lake?

Solution :

a) When t = 1.5, h(t) = ?

h(1.5) = -16(1.5)² + 64(1.5) + 80

h(1.5) = -16(2.25) + 96 + 80

h(1.5) = -36 + 96 + 80

h(1.5) = 140

b) Maximum height :

h(t) = -16[t² - 4t - 5]

h(t) = -16[t² - 2 ⋅ t ⋅ 2 + 2² - 2² - 5]

h(t) = -16[(t - 2)² - 4 - 5]

h(t) = -16[(t - 2)² - 9]

h(t) = -16(t - 2)² + 144

Vertex is at (2, 144). So, the maximum height is 144 ft.

c)  When h(t) = 0

0 = -16t² + 64t + 80.

-16[t² - 4t - 5] = 0

t² - 4t - 5 = 0

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

t = 5 and t = -1

After 5 seconds, the rocket will hit the ground.

Problem 6 :

A rock is thrown from the top of a tall building. The distance, in feet, between the rock and the ground t seconds after it is thrown is given by

d(t) = -16t² – 4t + 382

How long after the rock is thrown is it 370 feet from the ground

Solution :

d(t) = -16t² – 4t + 382

t = ?, when d(t) = 370

-16t² – 4t + 382 = 370

-16t² – 4t + 12 = 0

Divide by -4, we get

4t² + t - 3 = 0

(t + 1) (4t - 3) = 0

t = -1 and t = 3/4

After 0.75 seconds, the height of the rock will be at 370 ft.