Question 1107699: A student is to mark out a rectangular field so that its perimeter is 400 metres. Let x be the length of the field. Find a formula for the area, A, of the field in terms of x and by graphing the formula on a number plane, find the dimensions of the field which would make the area a maximum.
Found 3 solutions by stanbon, addingup, ikleyn:
Answer by stanbon(75887)   (Show Source):
You can put this solution on YOUR website! : A student is to mark out a rectangular field so that its perimeter is 400 metres. Let x be the length of the field. Find a formula for the area, A, of the field in terms of x and by graphing the formula on a number plane, find the dimensions of the field which would make the area a maximum.
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Draw a rectangle. Mark a side "x". Mark the side opposite "x".
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The sum of the 2 remaining sides is 400-2x
Each of those remaining sides is (400-2x)/2 = 200-x
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Area = x(200-x) = 200x - x^2
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Maximum occurs where 200-2x = 0
x = 100
200-x = 100
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Cheers,
Stan H.
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Answer by addingup(3677)  (Show Source):
You can put this solution on YOUR website! 2L + 2W = 400
simplifying:
L + W = 200
L = 200 - W
Equation for the area:
A = (200 - W) x W = 200W - W^2
So we have a negative quadratic, which means we'll have an upside down parabola where the vertex is the maximum.
Quadratic equation in the format y = ax2 + bx + c :
A = –W^2 + 200W
The vertex of the parabola is the point (h, k) where h = -b/2a:
h = -200/(2 x (-1)) = -200/-2 = 100
To find k, let's plug 100 for W:
k = -100^2 + 200(100) = -10,000 + 20,000 = 10,000
OK, so now we know that h (the maximizing width) is 100, and k (the maximum area) is 10,000. So the shape of our rectangle will be the square:
2L + 2W = 400
2(100) + 2(100) = 400
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NOTE: Of the rectangles (four sided figures with four right angles), the square will always give you the largest area.
And of all geometric figures, the one that will give you the biggest area is the circle. IF your figure was a circle, the area you would get would be:
A = C^2/4Pi where C is the perimeter
A = 400^2/12.56 12,739
Happy learning,
John
Answer by ikleyn(52792)  (Show Source):
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