# string of length 1 m is formed into a rectangle with one pair of opposite sides each x cm. calc. the value of x which will max are enclosed by strong

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We are given that the perimeter of the rectangle is 1m, and that one of the sides has length `x` , so let the length be `x` . Then since the perimeter `P` is equal to twice the width plus twice the length we have the width `w=1/2-x` since `P=2l+2w => 1=2(x)+2(1/2-x)` .

Then the area of the rectangle, `A=l*w` is `A=x(1/2-x)` .

We want to maximize the area. A function `f(x)` can have a local maximum (or minimum) only at its critical points -- the points where the first derivative is 0 or does not exist. Since this function is everywhere continuous and differentiable, we need only find the zero(s) of the first derivative.

`A(x)=x(1/2-x^2)=1/2x-x^2`

`A'(x)=1/2-2x`

Setting `A'(x)=0` yields:

`1/2-2x=0`

`2x=1/2`

`x=1/4`

Since `A'(x)>0` for `x<1/4` and `A'(x)>0` for `x>1/4` we know that `x=1/4` is a local maximum.

**So the largest area is achieved when the length=width=1/4.**

Note that this makes sense algebraically -- the function` ` A(x) is a quadratic whose graph is a prabola that opens down. The vertex, lying on the axis of symmetry, will be the maximum value. Here the vertex is at (1/4,1/16), so x=1/4 yields the maximal value.** **

**Sources:**

Let ABCD be the rectangle, and let AB=x. Then CD=x since opposite sides are congruent. The sum of the side lengths is 1, so if w is the width we have w+w+x+x=1 or 2w+2x=1. Then 2w=1-2x; dividing by 2 we get w= 1/2-x.

The question asked for the lengths in centimeters. I gave the answer in terms of the perimeter. 1/4 of a meter is 25 centimeters. Each side is 25 cm long.

and the answer according to my book is 25

thank you for your answer, but sorry why is the width 1/2-x?

*which will maximise the area enclosed by the string. sorry