Instruction: The given problem below has been solved for you. Your job is to check the answers for errors. If there are errors, do the following in the given table.  a. Review each step of the problem.  b. Determined whether the given step is correct or if there is an error.  c. If a particular step has an error, explain it and make necessary correction.  d. Explain the correction. Tata is working as a city Engineer. The city mayor asks him to propose an irregular shape playground design. Aside from this, the mayor wanted the playground to be covered with frog grass.  How much budget the Mayor should allocate to cover the entire park with frog grass? OTHER STEPS INCLUDED STEP 3: Region 1: Solve the shaded region between f(x)=-x^2/100+200 and g(x)=0 bounded from x=0 to x=200. Region 2: Solve the shaded region between f(x)=-x^2/100+200 and g(x)=0 bounded from x=0 to x=100. Region 3: Solve the shaded region between h(x)=x and g(x)=0 bounded from  x=0 to x=141.24. Step 8: Solve for the allocated budget if the entire playground will be covered by frog grass. Since the frog grass costs P110 per square meter, it follows that,  (Total area)(Price of the frog grass per square meter)=Total Budget Total Budget=(21 649.58)(110) Total Budget=2,381,453.8 Therefore, the mayor should allocate a total P2, 381,453.8 for the frog grass. Step 9: Conclusion With that said, the total area of the proposed park, which is 2.16 hectares, is within the minimum area the guide stated. And as the mayor suggested covering the playground with frog grass, the engineer had also calculated the estimated budget in covering the total area of frog grass, which resulted in the amount of P2, 381,453.8. Hence, a budget of P 2,381,453.8 must have allocated to construct a playground covered in frog grass

Advanced Engineering Mathematics
10th Edition
ISBN:9780470458365
Author:Erwin Kreyszig
Publisher:Erwin Kreyszig
Chapter2: Second-order Linear Odes
Section: Chapter Questions
Problem 1RQ
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Instruction: The given problem below has been solved for you. Your job is to check the answers for errors. If there are errors, do the following in the given table. 
a. Review each step of the problem. 
b. Determined whether the given step is correct or if there is an error. 
c. If a particular step has an error, explain it and make necessary correction. 
d. Explain the correction.

Tata is working as a city Engineer. The city mayor asks him to propose an irregular shape playground design. Aside from this, the mayor wanted the playground to be covered with frog grass.  How much budget the Mayor should allocate to cover the entire park with frog grass?

OTHER STEPS INCLUDED
STEP 3:
Region 1: Solve the shaded region between f(x)=-x^2/100+200 and g(x)=0
bounded from x=0 to x=200.
Region 2: Solve the shaded region between f(x)=-x^2/100+200 and g(x)=0
bounded from x=0 to x=100.
Region 3: Solve the shaded region between h(x)=x and g(x)=0 bounded from 
x=0 to x=141.24.

Step 8: Solve for the allocated budget if the entire playground will be covered by frog grass.
Since the frog grass costs P110 per square meter, it follows that, 

(Total area)(Price of the frog grass per square meter)=Total Budget

Total Budget=(21 649.58)(110)

Total Budget=2,381,453.8

Therefore, the mayor should allocate a total P2, 381,453.8 for the frog grass.

Step 9: Conclusion
With that said, the total area of the proposed park, which is 2.16 hectares, is within the minimum area the guide stated. And as the mayor suggested covering the playground with frog grass, the engineer had also calculated the estimated budget in covering the total area of frog grass, which resulted in the amount of P2, 381,453.8. Hence, a budget of P 2,381,453.8 must have allocated to construct a playground covered in frog grass. 

100
Step 5: Solve the
area of Region 2.
Area 2 =
[(upper function) – (lower function)] dx
100
x2
Area 2 =
- (0) dx
200
100
x²
Area 2 =
dx
200,
100
1
Area 2 =
x' dx
200
1
Area 2 = -
200
100
1
1003
Area 2 =
200
5,000|
Area 2
* 1666,67
Step 6: Solve the
141.24
area of Region 3.
Area 3 =
[(upper function)- (lower function)] dx
141.24
Area 3 =
[(x) – (0)] dx
141.24
Area 3 =
x dx
|141.24
Area 3 =
(141.24
- 0
Area 3 =
Area 3 8 6649.58
Step 7: Solve the
total area.
Total Area = Area 1+ Area 2 + Area 3
Total Area = 13,333.33 + 1666.67 + 6649.58
Total Area = 21,649.58
Therefore, the total area of the irregular-shaped playground is approximately 21,649.58
square meters, which is equal to approximately 2.16 hectares.
Transcribed Image Text:100 Step 5: Solve the area of Region 2. Area 2 = [(upper function) – (lower function)] dx 100 x2 Area 2 = - (0) dx 200 100 x² Area 2 = dx 200, 100 1 Area 2 = x' dx 200 1 Area 2 = - 200 100 1 1003 Area 2 = 200 5,000| Area 2 * 1666,67 Step 6: Solve the 141.24 area of Region 3. Area 3 = [(upper function)- (lower function)] dx 141.24 Area 3 = [(x) – (0)] dx 141.24 Area 3 = x dx |141.24 Area 3 = (141.24 - 0 Area 3 = Area 3 8 6649.58 Step 7: Solve the total area. Total Area = Area 1+ Area 2 + Area 3 Total Area = 13,333.33 + 1666.67 + 6649.58 Total Area = 21,649.58 Therefore, the total area of the irregular-shaped playground is approximately 21,649.58 square meters, which is equal to approximately 2.16 hectares.
Step 1: Proposed
layout of the
irregular-shape
playground and
(0, 200)
BUILDING
MALL
Step 2:
Sketch the graph
of the layout
design indicating
the points and
h(x)
functions
(100, 100)
PLAYGROUND
100
BUILDING
SUILDING
(-141.421, 0)
(11 421
|(0, 0)
Figure 2. Graph of the Irregular-Shaped Playground
HOSPITAL
Functions:
f(x) = -
+ 200
100
g(x) = 0
h(x) = x
200
Step 4: Solve the
area of Region 1.
Area 1 =
[(upper function) – (lower function)] dx
200
Area 1 =
- (0)
dx
200
200
Area 1 =
dx
200
200
1
Area 1 =
x? dx
200
1
200
Area 1 =
200 3
(2003
Area 1 = -
200
3
40,000
Area 1=
* 13,333.33
Transcribed Image Text:Step 1: Proposed layout of the irregular-shape playground and (0, 200) BUILDING MALL Step 2: Sketch the graph of the layout design indicating the points and h(x) functions (100, 100) PLAYGROUND 100 BUILDING SUILDING (-141.421, 0) (11 421 |(0, 0) Figure 2. Graph of the Irregular-Shaped Playground HOSPITAL Functions: f(x) = - + 200 100 g(x) = 0 h(x) = x 200 Step 4: Solve the area of Region 1. Area 1 = [(upper function) – (lower function)] dx 200 Area 1 = - (0) dx 200 200 Area 1 = dx 200 200 1 Area 1 = x? dx 200 1 200 Area 1 = 200 3 (2003 Area 1 = - 200 3 40,000 Area 1= * 13,333.33
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