Of all rectangles with a fixed perimeter p, which one has the maximum area? (give the dimensions in terms of p.)
Answers
Answer:
The square
Explanation:
Here are two answers, one without calculus and one without. Which one you need depends on what you're doing in class!
Without calculus.
Let x and y the length and breadth of the rectangle.
Then the perimeter is
2x + 2y = p.
The area is A = xy, and this is what we need to maximize.
Now
(p/2) - 2√A = x + y - 2√(xy) = ( √x - √y )² ≥ 0, since squares are never negative. Furthermore, the expression is equal to 0 only if x = y.
So...
2√A ≤ p / 2
=> A ≤ ( p / 4 )²
with equality only if x = y.
Since p is fixed, this means that A is maximized only if x = y.
That is, the rectangle of perimeter p with maximum area is the square.
With calculus
Let x and y the length and breadth of the rectangle.
Then the perimeter is 2x + 2y = p.
The area is A = xy, and this is what we need to maximize.
From the first equation, y = p/2 - x. Substituting this into the formula for A gives:
A = x ( p/2 - x ) = (p/2) x - x².
As x and y cannot be negative, we have x ∈ [ 0, p/2 ].
So the maximum value for A occurs at x = 0, x = p/2, or at a point where dA/dx = 0. We just need to find where dA/dx = 0 and check these candidates.
Now...
dA/dx = 0
=> p/2 - 2x = 0
=> 2x = p/2
=> x = p/4.
So the three candidates are x = 0, x = p/2 and x = p/4.
At x = 0, we have A = 0.
At x = p/2, we have A = 0.
At x = p/4, we have A = p/4 × ( p/2 - p/4 ) = p/4 × p/4 = p²/16.
As this last one gives the largest value for A, we conclude that A is maximized when x = p/4. In this case, y = p/2 - p/4 = p/4, too. So the rectangle is a square.
Given : rectangles with a fixed perimeter p
To Find : which one has the maximum area
Solution:
rectangles with a fixed perimeter p
Let say one side of rectangle = x
Then another side of rectangle = p/2 -x
Area of Rectangle A = x ( p/2 -x)
=>A = xp/2 - x²
dA/dx = p/2 - 2x
put dA/dx = 0
=> p/2 - 2x = 0
=> x = p/4
d²A/dx² = - 2 < 0
Hence x = p/4 will give maximum area
x = p/4
p/2 -x = p/2 - p/4 = p/4
Hence all sides = p/4
Hence Square
Of all rectangles with a fixed perimeter p , Square with side p/4 has maximum area
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