MATH 31
U5 L5 APPLIED MAXIMUM AND MINIMUM PROBLEMS
PROJECTILES AND NUMBER PROBLEMS
NAME
ANSWERS
_Page 1 of 4
MARK__________________/ 25
____________%
EXAMPLE 1: The parabolic path of an aircraft used to simulate weightlessness can be
represented by the quadratic equation h(t) = -10t2 + 300t + 9750 where h(t) is the
altitude of the aircraft, in metres, and t is the time, in seconds, since weightlessness was
achieved.
Find the time it takes to reach maximum altitude and the altitude reached
h′(x) = -20t + 300
-20t + 300 = 0
t = 15
h(15) = 12 000
The maximum altitude of 12 000 m
is reached after 15 seconds
EXAMPLE 2
Find two numbers whose sum is 16 and whose product is a maximum.
sum = x + y
product = xy
16 = x + y
y = 16 – x
P = x(16 – x) = 16x – x 2
P′ = 16 – 2x
P" = -2
16 – 2x = 0
negative means max
x=8
y = 16 – 8 = 8
The numbers and 8 and 8 and the maximum product is 64.
U5 L5ANS Projectiles and Number Problems
MATH 31
U5 L5 APPLIED MAXIMUM AND MINIMUM PROBLEMS
PROJECTILES AND NUMBER PROBLEMS
ANSWERS
_Page 2 of 4
NAME
EXAMPLE 3
The sum of two positive numbers is 4. If the sum of their cubes is a minimum, what
are these numbers?
x+ y=4
S = x3 + y3
y=4–x
S=x
y
=4–2=2
3
+ (4 – x)3
+ 6(4 – x)(– 1)
= 6x – 24 + 6x
= – 24
negative means max
S" = 6x
S′ = 3x 2 + 3(4 – x)2 (-1)
S′ = 3x 2 – 3(16 – 8x + x2)
S′ = 3x 2 – 48 + 24x – 3x2
S′ = 24x – 48
24x – 48 = 0
x=2
The numbers are 2 and 2
EXAMPLE 4
Find two consecutive natural (positive whole) numbers if the sums of the larger number
and four times the reciprocal of the smaller number is a minimum.
Let smaller number be x so the larger number is x + 1
𝟏
𝑺(𝒙) = (𝒙 + 𝟏) + (𝟒 × )
𝒙
𝑺(𝒙) = 𝒙 + 𝟏 + 𝟒𝒙−𝟏
𝑺′(𝒙) = 𝟏 − 𝟒𝒙−𝟐
𝟒
𝟎=𝟏− 𝟐
𝒙
𝟒
𝟏= 𝟐
𝒙
𝒙𝟐 = 𝟒
𝒙 = ±𝟐 but only +2 is natural
ASSIGNMENT QUESTIONS
U5 L5ANS Projectiles and Number Problems
𝑆"(𝒙) = 𝟖𝒙−𝟑
𝑆"(𝟐) =
=
𝟖
𝒙𝟑
𝟖
=𝟏
𝟐𝟑
Since P" is positive this
is a minimum value.
The numbers are 2 and 2 + 1 = 3
MATH 31
U5 L5 APPLIED MAXIMUM AND MINIMUM PROBLEMS
PROJECTILES AND NUMBER PROBLEMS
NAME
ANSWERS
_Page 3 of 4
1. Robbie observed that a golf ball followed a parabolic path for which the height is
approximated by the quadratic function h(t) = –4t2 + 14.4 t, where h(t) is the
height in metres and t is the time in seconds.
/7
a) Find the time the golf ball reaches its maximum height and the maximum height
reached by the golf ball.
h ′ (t)= – 8t + 14.4
Label the graph with
ordered pairs
– 8t + 14.4 = 0
( 1.8 , 12.96)
t = 1.8
h(1.8) = 12.96
The maximum height of 12.96 m is
reached after 1.8 seconds
height
b) What was the original height of the golf ball?
(0, 0)
The ball was on the ground at
height 0 metres
c) How long was the golf ball in the air?
The ball was in the air for 3.6 seconds – 4t2 + 14.4t = 0
– 4t(t – 3.6) = 0
time
t = 0 or t = 3.6
2. On a sloping street, Lori gives Michael, who is on a skateboard, an abrupt push
uphill.
After time, t, in seconds, the distance, s, in feet, that Michael travels from the
starting point is determined by the equation s(t) = 16t – 2t2 .
a) After how many seconds will he start to roll back down the street?
/4
s ′ (t) = 16 – 4t
16 – 4t = 0
t =4
It will roll back down the street after 4 seconds
b) Find the farthest distance Michael travels uphill before he starts to roll back
down.
s(4) = 16(4) – 2(4)2 = 32
Michael will travel a maximum distance of 32 feet
U5 L5ANS Projectiles and Number Problems
( 3.6 , 0 )
MATH 31
U5 L5 APPLIED MAXIMUM AND MINIMUM PROBLEMS
PROJECTILES AND NUMBER PROBLEMS
NAME
ANSWERS
_Page 4 of 4
3. A Frisbee is thrown straight up in the air from a position 2 m above the ground level.
Because of the wind pattern, the height, h, in metres, after time, t, in seconds, is given
by the formula h(t) = 2 + 6t – 2t2.
a) What is the maximum height the Frisbee will reach:
h ′ (t)= 6 – 4 t
6–4t =0
/4
t = 1.5
h(1.5) = 6.5
The maximum height of 6.5 m is reached after 1.5 seconds
b) If it is caught 2 m above the ground, how long will it have been in the air?
It will be in the air for 3 seconds
2
4. Find two numbers that differ by 18 and have a minimum product.
Difference = x – y
x – y =18
x = 18 + y
/5
product = xy
P = y(18+y)= 18y + y2
P’(x) = 18 + 2y
0 = 18 + 2y
y= –9
x = 18 + (– 9) = 9
The numbers are 9 and – 9
P" = 2 positive value means a minimum
5. Find two numbers whose sum is 48 and the sum of their squares is a minimum.
x + y = 48
S = x2 + y2
y = 48 – x
S = x 2 + (48 – x)2
y
= 48 – 24 = 24
/5
S ′ = 2x + 2(48 – x )(-1)
S ′ = 2x – 96 + 2x
S ′ = 4x – 96
4x – 96 = 0
x = 24
The numbers are 24 and 24 and the minimum is 1152
U5 L5ANS Projectiles and Number Problems
S"
= 4 positive