Asked by Anonymous
a rectangular garden is to be fenced along four sides and in the middle it is to be divided into 2 equal areas.if 300meters of fencing materials is available, find the maximum possible area that can be fenced.
Answers
Answered by
Reiny
make a sketch, so you will have 2 longs sides and 3 short sides parallel with the divider.
Let the long sides by y, and the short sides be x
3x + 2y = 300
y = (300 - 3x)/2 = 150 - 1.5x
Area = xy
= x(150-1.5x)
= 150x - 1.5x^2
= -1.5(x^2 - 100x + 2500 - 2500)
= -1.5(x-50)^2 + 3750
Max area = 3750 m^2
check: x = 50, y = 75
area = 3750, 3x+2y= 150+150 = 300
Let the long sides by y, and the short sides be x
3x + 2y = 300
y = (300 - 3x)/2 = 150 - 1.5x
Area = xy
= x(150-1.5x)
= 150x - 1.5x^2
= -1.5(x^2 - 100x + 2500 - 2500)
= -1.5(x-50)^2 + 3750
Max area = 3750 m^2
check: x = 50, y = 75
area = 3750, 3x+2y= 150+150 = 300
There are no AI answers yet. The ability to request AI answers is coming soon!
Submit Your Answer
We prioritize human answers over AI answers.
If you are human, and you can answer this question, please submit your answer.