MAXIMUM AND MINIMUM WORD PROBLEMS

Problem 1 :

The height H meters of a rocket t seconds after it is fired vertically upwards is given by

H(t)  =  100t - 5t², t > 0.

a) How long does it take for the rocket to reach its maximum height?

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

c) How long does it take for the rocket to fall back to earth?

Solution :

H(t)  =  100t - 5t², t > 0

(a)  Maximum value :

H(t)  =  -5t² + 100t

a  =  -5, b  =  100 and c  =  0

t  =  -b/2a

t  =  -100/2(-5)

t  =  10

So, after 10 seconds the rocket will reach the maximum height.

(b)  To find the maximum height, we apply t  =  10.

H(10)  =  -5(10)² + 100(10)

=  -5(100) + 1000

=  -500 + 1000

=  500

So, maximum height reached by the rocket is 500 meter.

c) The rocket falls when H(t)  =  0

100t - 5t²  =  0

-5t(20 - t)  =  0

t  =  0 and t  =  20

So, after 20 seconds the rocket will fall in the ground.

Problem 2 :

A manufacturer finds that the profit $P from assembling x bicycles per day is given by

P(x)  =  -x² + 50x - 200

a) How many bicycles should be assembled per day to maximize the profit?

b) What is the maximum profit?

c) What is the loss made if no bicycles are assembled in a day? Suggest why this loss would be made.

Solution :

(a)  Maximum value :

-x² + 50x - 200  =  0

a  =  -1,b  =  -50 and c  =  -200

x  =  -b/2a

x  =  -(-50)/2(-1)

x  =  25

So, 25 bicycles to be assembled to maximize the profit.

(b) To find maximum profit, we apply x  =  25

P(25)  =  -25² + 50(25) - 200

=  -625 + 1250 - 200

=  -825+1250

=  425

So, the maximum profit is $425.

(c) If no bicycles are sold, then x  =  0

P(x)  =  -x² + 50x - 200 

P(0)  =  -0² + 50(0) - 200 

P(0)  =  -200

If no bicycles are sold, then loss happened for $200.

Problem 3 :

The driver of a car travelling downhill on a road applied the brakes. The speed s of the car, t seconds after the brakes were applied, is given by

s(t) = -6t² + 12t +60 kmh^-1

a) How fast was the car travelling when the driver applied the brakes?

b) After how many seconds was the speed of the car 64.5 kmh^-1? Explain your answer.

c) After how many seconds did the car reach its maximum speed?

d) What was the maximum speed reached?

Solution :

(a)  At the time when break applies, t  =  0

Given :

s(t) = -6t² + 12t + 60

s(0)  =  60 km h^-1

(b)  Speed 64.5 kmh^-1

64.5 = -6t² + 12t + 60

6t² - 12t - 60 + 64.5  =  0

6t² - 12t + 4.5  =  0

By dividing the the equation by 3, we get

2t² - 4t + 1.5  =  0

20t² - 40t + 15  =  0

By dividing 5, we get

4t² - 8t + 3  =  0

(2t-3)(2t-1)  =  0

t  =  3/2 and t  =  1/2

Maximum speed :

a  =  -6, b  =  12 and c  =  60

x  =  -b/2a

x  =  -12/2(-6)

x  =  1

(c)  After 1 second the car reaches its maximum speed

(d)  Maximum speed :

s(1)  =  -6(1)² + 12(1) + 60

s(1)  =  -6+12+60

s(1)  =  -6+72

s(1)  =  66 kmh^-1

Maximum speed is 66 kmh^-1.

Problem 4 :

The hourly profit $P obtained from operating a fleet of n taxis is given by

P(n)  =  120n - 200 - 2n²

a) What number of taxis gives the maximum hourly profit?

b) What is the maximum hourly profit?

c) How much money is lost per hour if no taxis are on the road?

Solution :

(a)  P(n)  =  120n - 200 - 2n²

Maximum value at n = ?

-2n² + 120n - 200  =  0

a  =  -2, b  =  120 and c  =  -200

x  =  -b/2a

x  =  -120/2(-2)

x  =  120/4

x  =  30

So, 30 taxis will give maximum hourly profit.

(b)  The maximum hourly profit :

p(n)  =  -2n² + 120n - 200

p(30)  =  -2(30)² + 120(30) - 200

p(30)  =  -2(900) + 3600 - 200

p(30)  =  -1800 + 3600 - 200

p(30)  =  -2000 + 3600

p(30)  =  1600

So, the maximum hourly profit is $1600.

c) If no taxis are on the road, n = 0

p(0)  =  -2(0)² + 120(0) - 200

p(0)  =  - 200

So, loss of $200.

Problem 5 :

A stone was thrown from the top of a cliff 60 meters above sea level. The height of the stone above sea level t seconds after it was released is given by 

H(t) = -5t² + 20t +60 meters.

a) Find the time taken for the stone to reach its maximum height.

b) What was the maximum height above sea level reached by the stone?

c) How long did it take before the stone struck the water?

Solution :

a) Maximum value at x  =  ?

h(t)  =  -5t² + 20t +60

a  =  -5, b  =  20  and c  =  60

t  =  -b/2a

t  =  -20/2(-5)

t  =  2

(b)  At t = 2, it has reached maximum height

h(2)  =  -5(2)² + 20(2) +60

h(2)  =  -5(4) + 40 +60

h(2)  =  -20 + 100

h(2)  =  80

So, the maximum height reached by the stone is 80 meter.

(c)  at t  =  0

-5t²+20t+60  =  0

t²-4t-12  =  0

(t-6)(t+2)  =  0

t  =  6 and t  =  -2

So, it takes 6 seconds..

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