A farmer wishes to construct 3 adjacent enclosures alongside a river. Each enclosure is x feet wide and y feet long.?
2015-11-10 4:03 pm
A farmer wishes to construct 3 adjacent enclosures alongside a river. Each enclosure is x feet wide and y feet long. No fence is required along the river, so each enclosure is fenced along 3 sides. The total enclosure area of all 3 enclosures combined is to be 900 square feet. What is the least amount of fence required?
回答 (4)
2015-11-10 4:08 pm
✔ 最佳答案
A = 3xy = 900;F = 3y + 4x = 900/x + 4x;
dF/dx = 4 - 900/x^2,
which will be 0 when
x = sqrt(900/4) = 15.
Then y = 900/45 = 20, and
F = 60 + 60 = 120 feet.
2015-11-10 4:29 pm
m=4y+x
A=yx, A=900, so x=900/y then:
m=4y+900/y
m=(4y^2+900)/y
dm/dy=(8y^2-4y^2-900)/y^2
dm/dy=(4y^2-900)/y^2, since y>0
dm/dy=0 only when
4y^2=900
y^2=225
y=15
m(15)=(4y^2+900)/y=120ft
A=yx, A=900, so x=900/y then:
m=4y+900/y
m=(4y^2+900)/y
dm/dy=(8y^2-4y^2-900)/y^2
dm/dy=(4y^2-900)/y^2, since y>0
dm/dy=0 only when
4y^2=900
y^2=225
y=15
m(15)=(4y^2+900)/y=120ft
2015-11-10 5:20 pm
total area = x·3y = 900
y = 300/x
amount of fencing = 4x + 3y = 4x + 900/x
set first derivative to 0
4 - 900/x² = 0
x² = 225
x = 15
4x+900/x = 120 ft
y = 300/x
amount of fencing = 4x + 3y = 4x + 900/x
set first derivative to 0
4 - 900/x² = 0
x² = 225
x = 15
4x+900/x = 120 ft
2015-11-10 4:15 pm
if x is perpendicular to the river, then total fence = 4x + 3y
A = x(3y) = 3xy = 900
y = 300/x
P(x,y) = 4x + 3y
P(x) = 4x + 300/x
P ' = 4 - 300/x^2 = 0
300/x^2 = 4
x^2 = 300/4
x = 5 sqrt 3
P = 4x + 300/x = 4[5 sqrt 3] + 300/[5 sqrt 3] <<< simplify as needed
P = 20 sqrt 3 + 20sqrt 3 = 40sqrt 3 <<< answer
A = x(3y) = 3xy = 900
y = 300/x
P(x,y) = 4x + 3y
P(x) = 4x + 300/x
P ' = 4 - 300/x^2 = 0
300/x^2 = 4
x^2 = 300/4
x = 5 sqrt 3
P = 4x + 300/x = 4[5 sqrt 3] + 300/[5 sqrt 3] <<< simplify as needed
P = 20 sqrt 3 + 20sqrt 3 = 40sqrt 3 <<< answer
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