+97 A farmer has 1567 meters of fencing and wants to enclose a large rectangular area and divide it into 4 parallel pens. What are the dimensions of the entire rectangle such that the enclosed area is maximised? Round all values to TWO decimal places.
Length of parallels = ? [correct answer is 156.70 m]
Length of end sides = ? [correct answer is 391.75 m]
This is what i have done,
1567=2L+5W -> 1567/2=L+5w -> (1567/2)-L-5=w
A=L*W
A=L*((1567/2)-L-5)
Da/Dl= ??
+130081 Let x be the end side of the rectangular area....... and we have 2 of these. So, after using 2x for each end, we have (1567-2x)/5 for the length of one of the parallels, which forms the length of the other side of the rectangle.
So we have
A = L * W = (x)*(1567-2x)/5= (1/5)(1567x - 2x^2)
A' = (1/5)[1567 -4x)......setting this to 0, we have
1567 - 4x = 0
4x = 1567
x = 391.75 m ....and that's the length of the end
And [1567 -2(391.75)] / 5 = 156.7m .....and that's the length of one of the parallels (the other side of the rectangle)
+130081 Let x be the end side of the rectangular area....... and we have 2 of these. So, after using 2x for each end, we have (1567-2x)/5 for the length of one of the parallels, which forms the length of the other side of the rectangle.
So we have
A = L * W = (x)*(1567-2x)/5= (1/5)(1567x - 2x^2)
A' = (1/5)[1567 -4x)......setting this to 0, we have
1567 - 4x = 0
4x = 1567
x = 391.75 m ....and that's the length of the end
And [1567 -2(391.75)] / 5 = 156.7m .....and that's the length of one of the parallels (the other side of the rectangle)
