University:
University of Michigan
Course:
MATH 115 | Calculus I
Academic year:

2021
Views:

164

Pages:

2
Author:

viejoteatroov

Math 115 / Exam 2 (March 30, 2021) page 6 4. [10 points] A landscaper is designing a rectangular garden surrounding a circular fountain in the middle. • The diameter of the fountain is 2 meters. • The distance from the fountain to the eastern and western edges of the garden is a meters. • The distance from the fountain to the northern and southern edges of the garden is b meters. • The part of the garden outside of the circular fountain will be covered with exactly 300 square meters of grass. b a a 2 b a. [4 points] Write a formula for b in terms of a. Solution: Observe that the radius of the fountain is 1 meter, so that its area is π · 12 = π. The length of the rectangle is 2 + 2a, and the width is 2 + 2b, so the area of the rectangle is (2 + 2a)(2 + 2b). The area of the part of the garden outside the circular fountain is the difference between these two numbers we have found: (2 + 2a)(2 + 2b) − π. We are told that this area is equal to 300 meters, and so we have an equation that we can use to solve for b: (2 + 2a)(2 + 2b) − π = 300 300 + π 2 + 2b = 2 + 2a 300 + π − 4a − 4 1 300 + π −2 = . b= 2 2 + 2a 4a − 4 b. [2 points] Write a formula for the function P (a) which gives the rectangular perimeter of the garden in terms of a only. Solution: The rectangular perimeter of the garden is equal to 2(2a + 2) + 2(2b + 2). We may use our equation 2 + 2b = 300+π 2+2a to write this in terms of a: P (a) = 2(2a + 2) + 2 300 + π 2 + 2a . Math 115 / Exam 2 (March 30, 2021) page 7 c. [4 points] In the context of this problem, what is the domain of P (a)? Solution: The domain of P (a) is constrained by the following two facts: • a cannot be less than 0. • b cannot be less than 0. From the first fact, we see that the domain of P (a) must begin with “[0.” Let us express the second fact as a statement about a. Using the formula b = 12 300+π 2+2a − 2 , we see that the second fact can be expressed as: 1 300 + π 0≤ −2 2 2 + 2a 300 + π 2≤ 2 + 2a 4 + 4a ≤ 300 + π 300 + π − 4 a≤ . 4 ].” Putting these together, Therefore, we see that the domain of P (a) must end with “ 300+π−4 4 we conclude that the domain of P (a) is [0, 300+π−4 ]. 4