
Suppose you have $ 200 $ feet of fencing to enclose a rectangular plot. How do you determine dimensions of the plot to enclose the maximum area possible?
Answer
531k+ views
Hint: The maximum area for a rectangular figure (for a fixed perimeter) is obtained when the figure is a square. Which means that the four sides should be the same in length. Also recheck this theory once by considering them as length and breadth of different units and then comparing them with maximized area.
Complete step-by-step answer:
We are given a fence of $ 200 $ feet to enclose a rectangular plot.
Since we know that the maximum area for a rectangular figure can be obtained when the figure has all sides equal or when it’s a square.
So, according to that theory if it’s a square then its length of sides should be $ \dfrac{{200}}{4} = 50 $ feet each side.
But, suppose if we don’t remember this fact then let’s prove it by solving.
Let the length of the rectangle be $ x $ feet. So the width would be $ (100 - x) $ feet as its given that perimeter $ \left( {2 \times length + 2 \times width} \right) = 200 $ feet
$
2\left( {l + b} \right) = 200 \\
l + b = 100 \;
$
Let, $ f\left( x \right) $ be a function for are of plot for a length $ x $ then : $ f\left( x \right) = l \times b = x \times \left( {100 - x} \right) = 100x - {x^2} $
And we obtained a simple quadratic equation with a maximum value at that point where its derivative would be $ 0 $ .
Differentiating with $ f\left( x \right) = l \times b = x \times \left( {100 - x} \right) = 100x - {x^2} $ respect to $ x $ and we get:
$ f'\left( x \right) = 100 - 2x $
Now, for its maximum value:
$
f'\left( x \right) = 0 \\
100 - 2x = 0 \;
$
On further solving, we get:
$
100 - 2x = 0 \\
2x = 100 \\
x = 50 \;
$
Therefore length $ \left( l \right) = x = 50 $ feet and breadth $ \left( b \right) = \left( {100 - x} \right) = 100 - 50 = 50 $ feet
And it proves that the above mentioned theory/fact was correct.
Hence, length of $ 50 $ feet and breadth of $ 50 $ feet is required as dimensions for the rectangular plot of $ 200 $ feet to enclose maximum area.
So, the correct answer is “length of $ 50 $ feet and breadth of $ 50 $ feet”.
Note: There can be errors in differentiating.
There can be a misunderstanding in understanding the question of whether an area is given or a perimeter for the figure.
Always check the solution once by putting the value, before coming into a conclusion.
Complete step-by-step answer:
We are given a fence of $ 200 $ feet to enclose a rectangular plot.
Since we know that the maximum area for a rectangular figure can be obtained when the figure has all sides equal or when it’s a square.
So, according to that theory if it’s a square then its length of sides should be $ \dfrac{{200}}{4} = 50 $ feet each side.
But, suppose if we don’t remember this fact then let’s prove it by solving.
Let the length of the rectangle be $ x $ feet. So the width would be $ (100 - x) $ feet as its given that perimeter $ \left( {2 \times length + 2 \times width} \right) = 200 $ feet
$
2\left( {l + b} \right) = 200 \\
l + b = 100 \;
$
Let, $ f\left( x \right) $ be a function for are of plot for a length $ x $ then : $ f\left( x \right) = l \times b = x \times \left( {100 - x} \right) = 100x - {x^2} $
And we obtained a simple quadratic equation with a maximum value at that point where its derivative would be $ 0 $ .
Differentiating with $ f\left( x \right) = l \times b = x \times \left( {100 - x} \right) = 100x - {x^2} $ respect to $ x $ and we get:
$ f'\left( x \right) = 100 - 2x $
Now, for its maximum value:
$
f'\left( x \right) = 0 \\
100 - 2x = 0 \;
$
On further solving, we get:
$
100 - 2x = 0 \\
2x = 100 \\
x = 50 \;
$
Therefore length $ \left( l \right) = x = 50 $ feet and breadth $ \left( b \right) = \left( {100 - x} \right) = 100 - 50 = 50 $ feet
And it proves that the above mentioned theory/fact was correct.
Hence, length of $ 50 $ feet and breadth of $ 50 $ feet is required as dimensions for the rectangular plot of $ 200 $ feet to enclose maximum area.
So, the correct answer is “length of $ 50 $ feet and breadth of $ 50 $ feet”.
Note: There can be errors in differentiating.
There can be a misunderstanding in understanding the question of whether an area is given or a perimeter for the figure.
Always check the solution once by putting the value, before coming into a conclusion.
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