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sienna has 800 yards of fencing to enclose a rectrangular area. find the dimensions of the rectangle that maximise the enclosed area
Area can be maximized by fencing a square of side 200 yards.
Explanation:
Given perimeter of a rectangle, square has the maximum area (proof given below).
Let x be one of the side and a be te perimeter then the other side would be a/2 - x
and area would be x (a/2 - x). The function will be zero when first derivative of the function is equal to zero and second derivative is negative,
As first derivative is − 2 x + a/2 and this will be zero, when − 2 x + a/2 = 0 or x = a/4. Note that second derivative is − 2. Then two sides will be a/4 each that the it would be square.
Hence if perimeter is 800 yards and it is a square, one side would be 800/4 = 200 yards.
Hence area can be maximized by fencing a square of side 200 yards.

Respuesta :

As per the area of rectangle, the dimensions of the rectangle that maximize the enclosed area is 200 yard.

Area of rectangle:

The standard formula to calculate the are of the rectangle is.

A = l x b

l refers the length

b refers the breadth

Given,

Sienna has 800 yards of fencing to enclose a rectangular area.

Area can be maximized by fencing a square of side 200 yards.

Here we need to find the dimensions of the rectangle that maximize the enclosed area.

Let us consider  x be one of the side and a be the perimeter then the other side would be written as,

=> a/2 - x

And the and area of the rectangle would be

=> x (a/2 - x).

Then the function will be zero when first derivative of the function is equal to zero and second derivative is negative,

So, as per the first derivative is written as,

=> − 2 x + a/2

And this will be zero, when the equation,

=> − 2 x + a/2 = 0 or x = a/4.

Here we have to note that second derivative is − 2.

And then the two sides will be a/4 each that the it would be square.

Therefore, if perimeter is 800 yards and it is a square, one side would be 800/4 = 200 yards.

To know more about Area of rectangle here.

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