Asked by Nahel
A rectangular lot is to be bounded by a fence on three sides and by a wall on the fourth side. Two kinds of fencing will be used with heavy duty fencing selling for $4 a foot on the side opposite the wall. The two remaining sides will use standard fencing selling for $3 a foot. What are the dimensions of the rectangular plot of greatest area that can be fenced in at a cost of $6600?
THANK YOU!!!
THANK YOU!!!
Answers
Answered by
Damon
cost = 6 b + 4 L = 6600
area = b L
so
4 L = 6600 - 6 b
L = 1650 - 1.5 b
area = y = 1650 b - 1.5 b^2
that is a parabola opening down, the maximum will be at the vertex
find the vertex
1.5 x^2 -1650 x = -y
x^2 - 1100 x = -2 y/3
x^2 - 1100 x + 302500 = -2 y/3 + 302500
(x-550)^2 = -2 y/3 + 302500
x = b = 550
so L = 1650 - 1.5 (550)
L = 825
area = b L
so
4 L = 6600 - 6 b
L = 1650 - 1.5 b
area = y = 1650 b - 1.5 b^2
that is a parabola opening down, the maximum will be at the vertex
find the vertex
1.5 x^2 -1650 x = -y
x^2 - 1100 x = -2 y/3
x^2 - 1100 x + 302500 = -2 y/3 + 302500
(x-550)^2 = -2 y/3 + 302500
x = b = 550
so L = 1650 - 1.5 (550)
L = 825
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