Optimizing with Constraints A rectangular garden is to be fenced using the wall of a house as one of side of the garden. The garden should have an area of 40 m². Determine the minimum perimeter and dimensions of the garden in each case: a) The dimensions must be whole numbers of metres. b) The dimensions can be decimals. 200
Optimizing with Constraints A rectangular garden is to be fenced using the wall of a house as one of side of the garden. The garden should have an area of 40 m². Determine the minimum perimeter and dimensions of the garden in each case: a) The dimensions must be whole numbers of metres. b) The dimensions can be decimals. 200
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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![Optimizing with Constraints
A rectangular garden is to be fenced using the
wall of a house as one of side of the garden. The
garden should have an area of 40 m². Determine
the minimum perimeter and dimensions of the
garden in each case:
a) The dimensions must be whole numbers of
metres.
b) The dimensions can be decimals.
Area=40 sqm, means I*w=40, but
P=1+2w
1*w=40 |+w+w=
1*40=40 1+40+40=81
2*20=40 2+20+20=42
4*10=40 4+10+10=24
5*8=40 5+8+8=21
8*5=40 8+5+5=18
¹10*4-40 10+4+4=18
20*2=40 20+2+2=24
40*1=40 40+1+1=42
10
Jo
y
x - l
I*w=40
|+w+w=
from part a minimum
perimeter in 2 situations.
length is 8 m and 10 m
average is (8+10)/2=9
9*4.444=40
P=9+4.4+4.4=17.8.
کیا۔
-20
(8.944, 17.889)](/v2/_next/image?url=https%3A%2F%2Fcontent.bartleby.com%2Fqna-images%2Fquestion%2F1cb0f8c5-7d0f-47d1-8a57-d99c12df3a50%2F90ad4043-325d-4291-af06-c4b7ac2a4c1a%2Fh8y7oeo_processed.jpeg&w=3840&q=75)
Transcribed Image Text:Optimizing with Constraints
A rectangular garden is to be fenced using the
wall of a house as one of side of the garden. The
garden should have an area of 40 m². Determine
the minimum perimeter and dimensions of the
garden in each case:
a) The dimensions must be whole numbers of
metres.
b) The dimensions can be decimals.
Area=40 sqm, means I*w=40, but
P=1+2w
1*w=40 |+w+w=
1*40=40 1+40+40=81
2*20=40 2+20+20=42
4*10=40 4+10+10=24
5*8=40 5+8+8=21
8*5=40 8+5+5=18
¹10*4-40 10+4+4=18
20*2=40 20+2+2=24
40*1=40 40+1+1=42
10
Jo
y
x - l
I*w=40
|+w+w=
from part a minimum
perimeter in 2 situations.
length is 8 m and 10 m
average is (8+10)/2=9
9*4.444=40
P=9+4.4+4.4=17.8.
کیا۔
-20
(8.944, 17.889)
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