Asked by Anonymous
A rectangular pen with one side along a river will be bounded by a fence on the other three sites. If 600m of fence are available, what the dimensions of the rectangle should be in order to enclose maximum area and how much this area is?
Answers
Answered by
Damon
x along river and y
600 = x + 2y so x = 600-2y
A = x y
A = (600-2y)y = 600y - 2y^2
I assume you do not know calculus so complete square to find vertex of parabola.
y^2-300 y = -A/2
y^2 -300 y+(300/2)^2 = -A/2 + 150^2
(y-150)^2 = -(1/2) (A-45,000)
vertex at y = 150 and A = 45,000
so
y = 150
x = 300
A = 45,000
600 = x + 2y so x = 600-2y
A = x y
A = (600-2y)y = 600y - 2y^2
I assume you do not know calculus so complete square to find vertex of parabola.
y^2-300 y = -A/2
y^2 -300 y+(300/2)^2 = -A/2 + 150^2
(y-150)^2 = -(1/2) (A-45,000)
vertex at y = 150 and A = 45,000
so
y = 150
x = 300
A = 45,000
Answered by
Reiny
length of side parallel to river --- y
length of the other two equal sides --- x
2x + y = 600 ---> y = 600-2x
area = xy
= x(600-2x)
= -2x^2 + 600x
this is a downwards opening parabola.
Find the vertex by whatever means you were taught to find the vertex, and extract your answer from that.
length of the other two equal sides --- x
2x + y = 600 ---> y = 600-2x
area = xy
= x(600-2x)
= -2x^2 + 600x
this is a downwards opening parabola.
Find the vertex by whatever means you were taught to find the vertex, and extract your answer from that.
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