Math 161 - Solutions To Sample Exam 2 Problems Worksheet Page 6
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8Solutions To Sample Exam 2 Problems – Math 161
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First, let x denote the radius of the semicircular portion of the
garden, and let y denote the length of one of the rectangular sides
of the garden (see diagram to the right). Therefore, we need 2y
feet of cheap fence, at a cost of $2 a foot, giving us the following:
x
Cost of Cheap Fence = 2y(2) = 4y dollars
y
The expensive fence forms a semicircle of radius x feet, so we need
(1/2)(2πx) = πx feet of expensive fence, at a cost of $5 a foot,
giving us the following:
Cost of Expensive Fence = (πx)(5) = 5πx dollars
Thus, the combined total cost of the fencing is given by 4y + 5πx dollars, and since we have only $200 to
spend, this means that 4y + 5πx = 200 (see “Given:” in the box below). To get the total area of the garden,
we have to add the area of the semicircle and rectangle, which gives the following:
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2
Total Area of Garden =
πx
+ 2xy
2
Therefore, we summarize the setup portion of the problem as follows:
Given:
4y + 5πx = 200
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2
Find:
x and y that maximize A =
πx
+ 2xy
2
REWRITE A AS A FUNCTION OF ONLY ONE VARIABLE:
Using our given information, we first solve for y:
4y + 5πx = 200
=⇒
4y = 200
5πx
200
5πx
=⇒
y =
4
5π
=⇒
y = 50
x.
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Substituting y into our formula for A and simplifying, we get
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1
5π
2
2
A(x) =
πx
+ 2xy =
πx
+ 2x · 50
x
2
2
4
π
5π
2
2
=
x
+ 100x
x
2
2
2
= 100x
2πx
.
2
Therefore, A(x) = 100x
2πx
is our simplified formula for the total area of the garden. To verify that a
maximum occurs at x ≈ 7.96, we make a sign chart:
LOCATE THE MAXIMUM VALUE OF A(x) :
First, we need to find the critical points of A(x). We have
A (x) = 0
=⇒
100
4πx = 0
=⇒
100 = 4πx
25
=⇒
x =
≈ 7.96,
π
so the only critical point occurs at x ≈ 7.96. The sign chart below confirms that a maximum value occurs at
this critical point:
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