Asked by Mya
A rectangular field is to be enclosed by a fence and divided into two smaller plots by a fence parallel to one of the side. Find the dimensions of the largest such field if 1200 m of fencing material is available. What is the area of this field and what are the dimensions that will give the largest area?
Answers
Answered by
Steve
Draw a diagram. There will be 3 strips of one length, and 2 strips of the other. If we call them x and y, then
3x+2y=1200
The area is
A = xy=x(1200-3x)/2 = 600x-3/2 x^2
The maximum area occurs at the vertex of the parabola, at
x = -b/2a = 200
So, there will be 3 lengths of 200
and 2 lengths of 300
The maximum area is 60,000 m^2
As with <u>all</u> such problems, the maximum area is achieved when the fencing is divided equally between lengths and widths, no matter how many of each there are.
In this case, 1200/2 = 600
since there are 3 lengths (x), each is 200
There are 2 widths (y), each is 300
3x+2y=1200
The area is
A = xy=x(1200-3x)/2 = 600x-3/2 x^2
The maximum area occurs at the vertex of the parabola, at
x = -b/2a = 200
So, there will be 3 lengths of 200
and 2 lengths of 300
The maximum area is 60,000 m^2
As with <u>all</u> such problems, the maximum area is achieved when the fencing is divided equally between lengths and widths, no matter how many of each there are.
In this case, 1200/2 = 600
since there are 3 lengths (x), each is 200
There are 2 widths (y), each is 300
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