The dimension of the rectangle of maximum area that you can enclose with the string is 12.5 x 12.5
n let x = the length of one side of the rectangle and y = the length of the other side of the rectangle.
The area is:
A = xy
and the perimetere is the string of length 50 cm
50 = 2(x + y) Solve this in terms of y. Divide both sides by 2.
25 = x + y By Subtract x from both sides.
25-x = y Now substitute this into the equation for the area.
A = x (25 - x) Simplify.
A = 25x - x²
the general form of quadratic equation: ax² + bx + c
x is given by -b/2a
Here the equation is A = -x² + 25x
, a = -1, b = 25, and c = 0
so we can find the x-coordinate of the vertex by -25/2(-1) = 12.5
So the length (x) of the rectangle must be 12.5 cm to get the maximum area.
But what about the width (y)?
Since the perimeter is 50 cm and this is twice the (length + width), the (length + width) is 25 cm, so the width is 25 cm - the length.
x + y = 25
12.5 + y = 25
y = 12.5
So, we end up with x (the length) = 12.5 cm and y (the width) = 12.5 cm.
And this, of course, is a square whose sides are 12.5 cm
If the perimeter were k cm,
2(x + y) = k
x + y = k/2
x + x = k/2 (as x = y)
x = k/4
x = k/4, y = k/4
Therefore, the dimension of the rectangle of maximum area that you can enclose with the string is 12.5 x 12.5
To learn more about perimeter refer here
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