Asked by Josh
A rectangular lot is to be fenced using 80m of fencing. What is the maximum area and what are the dimensions of the lot?
Answers
Answered by
Anonymous
P=Perimeter
A=Area
P=2a+2b
80=2a+2b Divide both sides with 2
40=a+b
40-b=a
a=40-b
A=a*b
A=(40-b)*b
A=40b-b^2
dA/dB=40-2b
Function has maximum where first derivation=0
and second derivation<0
First derivation:
dA/dB=40-2b=0
40=2b Divide booths sides with 2
20=b
b=20m
a=40-b=40-20=20m
Second derivation= -2
For a=20m and b=20m rectangle with maximum area is a square.
A(max)=20*20=400m^2
A=Area
P=2a+2b
80=2a+2b Divide both sides with 2
40=a+b
40-b=a
a=40-b
A=a*b
A=(40-b)*b
A=40b-b^2
dA/dB=40-2b
Function has maximum where first derivation=0
and second derivation<0
First derivation:
dA/dB=40-2b=0
40=2b Divide booths sides with 2
20=b
b=20m
a=40-b=40-20=20m
Second derivation= -2
For a=20m and b=20m rectangle with maximum area is a square.
A(max)=20*20=400m^2
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