Download Section 3.3B Word Problems

Section 3.3B Word Problems: Maximizing (Finding the Maximum using the Vertex)
Example1:
At a concert, organizers are roping off a rectangular area for sound equipment. There is 160m
of fencing available to create the perimeter. What dimensions will give the maximum area, and
what is the maximum area?
Steps:
1) Write an equation for the perimeter, and write an equation for the area for the rectangle.
2) Use the two equations to create a quadratic function in standard form.
3) Change the quadratic function into vertex-graphing form.
4) Identify the maximum area, and then the dimensions for the maximum area.
y
1) If we let x be the width and y be the length, A is the area.
Perimeter = 2x + 2y with x and y being the width and length
So 160 = 2x + 2y (Solve for y to create a function.)
160 – 2x = 2y
80 – x = y
y = 80 – x
x
Sound
Equipment
y
Area = xy
A = xy
2) Now combine the 2 equation from part 1) to find an equation to maximize the area.
y = 80 – x
A = xy
A = x(80 – x)
A = 80x – x2
A = -x2 + 80x
b
formula.
2a
Since all we need is the vertex, the simpler and quicker way is to use the formula to find the
vertex. A = -x2 + 80x
a = -1, b = 80
b
80
(x, A)
p=−
=−
= 40 So the vertex is (40, ??)
2a
2(−1)
Substitute 40 in for x and we get:
A = -(40)2 + 80(40) = -1600 + 3200 = 1600
Our vertex is (40, 1600)
3) Now complete the square to change into vertex-graphing form OR use p = −
4) Using the vertex, our maximum area is 1600m2 and one dimension of the box is 40m.
Since A = xy 1600=40y so y = 40 The box is 40m by 40m.
x
Graph of the equation: A = -x2 + 80x
A
1800
1600
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1000
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200
x
0
10
20
30
40
50
60
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100
Example2:
A sporting goods store sells basketball shorts for $8. At this price their weekly sales are
approximately 100 items. Research says that for every $2 increase in price, the manager can
expect the store to sell five fewer pairs of shorts. Determine the maximum revenue the
manager can expect based on these estimates. What selling price will give that maximum
revenue, and how many shorts will be sold?
Let x = # of $2 price increases
Let R = the revenue
Based on no changes, the revenue would have been R = (price)(shorts_sold) = (8)(100).
With the expected changes:
The new price = 8 + 2x
{for every increase, price goes up by $2}
The shorts sold = 100 – 5x {for every increase, number of shorts sold goes down by 5}
Formula:
R = (8 +2x)(100 – 5x)
To get the maximum revenue, rewrite the above as a quadratic function and find the vertex
(maximum values).
R = 800 – 40x + 200x – 10x2
R = -10x2 + 160x + 800
Use the formula p = −
b
to help find the vertex.
2a
R = -10x2 + 160x + 800
a = -10, b = 160
p=−
b
160
=−
= 8 So the vertex is (8, ???)
2a
2(−10)
Substitute 8 in for x in the equation R = -10x2 + 160x + 800 to find the other part of the vertex.
R = -10(8)2 + 160(8) + 800
R = -640 + 1280 +800
R = 1440
So the vertex is (8, 1440)
This means there are 8 increases of $2 with a maximum revenue of $1440.
The new price will be 8 + 2x = 8 + 2(8) = $24
The number of shorts sold will be 100 – 5x = 100 – 5(8) = 60 shorts
Graph:
R
1600
1400
1200
1000
800
600
400
200
0
-24 -20 -16 -12
-8
-4
0
4
8
12
16
20
24
x
Word Problem questions for students to try:
#1. Cathy has 48m of fencing to make a rectangular pen for her pet. What is the maximum
area? Answer: 144m2
Let x = width
Let 24-x = length
(P = 2L + 2W
48 = 2L + 2W
24 = L + W L = 24 – W)
A = x(24-x)
A = 24x – x2
A = -x2 + 24x
Find vertex (p, q) to find maximum Area.
b
24
p=
−
=
−
=
12
2a
2(−1)
q = -(12)2 + 24(12)
q = -144 + 288
q = 144
So the maximum area = 144m2
#2. The Summer Theatre charges $10 per ticket and it has had a full house of 500 nightly. The
manager estimates that the ticket sales would decrease by 50 for each $2 increase in ticket
cost. What is the most profitable price to charge for each ticket? At this new price what is the
maximum revenue the Summer Theater can expect to generate?
Answer: ticket price = $15 max revenue = $5625
Let x = number of $2 increases
Let R = revenue
R = price x numberoftickets
R = (10 + 2x)(500 – 50x)
R = 5000 – 500x + 1000x – 100x2
R = -100x2 + 500x + 5000
Find vertex (p, q) to find the Max revenue.
b
500
p=
−
=
−
=
2.5
2a
2(−100)
q = -100(2.5)2 + 500(2.5) + 5000
q = 5625
The Max revenue is $5625 .
The price is 10 + 2(2.5) = $15 per ticket
#3. A management firm has determined that 60 apartments in a complex can be rented if the
monthly rent is $900, and that for each $50 increase in the rent, three tenants are lost with
little chance of being replaced. What rent should be charged to maximize revenue? What is the
maximum revenue?
Answer: $950 per month for rent max revenue = $54,150
Let x = number of $50 increases
Let R = revenue
R = monthly rent x numberofapartment
R = (900 + 50x)(60 – 3x)
R = 54000 – 2700x + 3000x – 150x2
R = -150x2 + 300x + 54000
b
300
p=
−
=
−
=
1
2a
2(−150)
q = -150(1)2 + 300(1) + 54000
q = 54150
The Max revenue is $54150 .
The monthly rent is 900 + 50(1) = $950 per apartment
Textbook Assignment: Page 194-196 #14, 18, 19