Question

Find the dimensions of a rectangle with a perimeter of 100 m whose area is as large as possible.

Short answer

25m is the length and 25m is the width.

Step-by-step solution

Given Data

1)Perimeter of the rectangle is 100 m.

2)Area is as large as possible.

Determination of the dimensions of the rectangle

Let length of the rectangle be x and width be y.

Now, \({\rm{Perimeter}} = 2\left( {x + y} \right)\)

Therefore, \(100 = 2\left( {x + y} \right)\)

Divide both sides by 2.

\(50 = x + y\)

Eliminating y from above equation,

\(y = 50 - x\)………..(1)

\(Area = xy\)

Substitute y from equation (1):

\(A\left( x \right) = x\left( {50 - x} \right)\)

\(A\left( x \right) = 50x - {x^2}\)

\(A'\left( x \right) = 50 - 2{x^{}}\)

Now, substitute \(A'\left( x \right) = 0\)as:

\(\begin{aligned}{c}50 - 2x &= 0\\2x &= 50\\x &= 25\end{aligned}\)

Put \(x = 25\)in equation (1) as:

\(y = 50 - x\)

\(\begin{aligned}{l}y &= 50 - x\\y &= 25\end{aligned}\)

Thus, the required dimensions of the rectangle is 25m in length and 25m in width.