Unformatted text preview: Solutions to Quiz 8B
www.math.uﬂ.edu/˜harringt
November 26, 2006
1. (3 pts.) 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.
have this image: So we 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. (1 pts.) Express the given equation in logarithmic form: 54 = 625.
Solution: log5 625 = 4
3. (1 pt.) Use the laws of logarithms to simplify the given expression: log x3 (x + 1)
Solution: log x3 (x + 1) = log x3 + log(x + 1) = 3 log x + log(x + 1) 1 ...
View
Full Document
 Spring '08
 Smith
 Math, Calculus, 1 pt, Logarithm, 1 pts, log x3

Click to edit the document details