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Unformatted text preview: Solutions to Quiz 11A
www.math.uﬂ.edu/˜harringt
April 11, 2008
1. Mrs. Zerta wants to start a garden next to her house. She only has 200 yards of
fencing. If fencing is NOT required along side of her house, what are the dimensions
of the largest area that she can enclose? What is this area?
Solution: We will let x denote the width and y denote the length. So we have
this image: So we have that the perimeter is P = 2x + y = 200. Also recall that the area
is: A = xy . Thus, we will solve for y using the perimter. So, y = 200 − 2x.
Next, we will substitute y = 200 − 2x back in the formula for area. Hence,
A = x(200 − 2x) = 200x − 2x2 . This is the equation we want to maximize. Also note
that, x ≥ 0, and y = 200 − 2x ≥ 0. So, 0 ≤ x ≤ 100. Therefore, we want to ﬁnd
the absolute max for A = 200x − 2x2 on the interval on [0, 100]. To ﬁnd the critcial
number we set A = 0 so A (x) = 200 − 4x = 0 implies that x = 50. (Note: if we
check the enpoints we just get A(0) = 0 and A(100) = 0.) So y = 200 − 2(50) = 100.
Thus the dimensions are 50 yards by 100 yards. The area is A = 50 ∗ 100 = 5000
yrds2 .
2. Supose that g (x) = ex and g (0) = 0, then what is g ?
So g (x) = ex + c. Since g (0) = 0 then e0 + c = 1 + c = 0 implies c = −1. Thus,
g (x) = ex − 1
1
3. Suppose that f (x) = x2 − 5x and f (1) = 0, then what is f ? So f (x) = 3 x3 − 5 x2 + c.
2
1
5
13
Since f (1) = 0, then 3 − 2 + c = 0 implies c = 6 . Thus, 1
5
13
f (x) = x3 − x2 +
3
2
6 1 ...
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This note was uploaded on 07/10/2011 for the course MAC 2311 taught by Professor All during the Fall '08 term at University of Florida.
 Fall '08
 ALL
 Math, Calculus, Geometry

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