Question

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

Short answer

The dimensions of a rectangle for the maximum area are \[x = y = 25{\rm{ meters}}\].

Step-by-step solution

Optimization of a Function

TheOptimization of a Function\[f\left( x \right)\]for any real value of\[c\]can be given as:

\[\begin{aligned}{l}f\left( x \right)\left| {_{x = c} \to \max \to \left\{ \begin{aligned}{l}{\rm{if }}f'\left( x \right) > 0\,\,\,\,\,\,\forall x < c\\{\rm{if }}f'\left( x \right) < 0\,\,\,\,\,\,\forall x > c\end{aligned} \right.} \right.\\{\rm{and}}\\f\left( x \right)\left| {_{x = c} \to \min \to \left\{ \begin{aligned}{l}{\rm{if }}f'\left( x \right) < 0\,\,\,\,\,\,\forall x < c\\{\rm{if }}f'\left( x \right) > 0\,\,\,\,\,\,\forall x > c\end{aligned} \right.} \right.\end{aligned}\].

Optimizing the function according to a given condition.

Let the dimensions be \[x{\rm{ and }}y\]. Then, the perimeter will give:

\[\begin{aligned}{c}P = 100\\2x + 2y = 100\\x + y = 50\\y = 50 - x\end{aligned}\].

Here, we have the area as:

\[\begin{aligned}{c}A\left( x \right) = x\left( {50 - x} \right)\\ = 50x - {x^2}\end{aligned}\]

For maximization, we have:

\[\begin{aligned}{c}A'\left( x \right) = \frac{d}{{dx}}\left( {50x - {x^2}} \right)\\\frac{d}{{dx}}\left( {50x - {x^2}} \right) = 0\\50 - 2x = 0\\x = 25\end{aligned}\]

Now, we have:

\[\begin{aligned}{l}{\rm{for }}0 < x < 25 \to \,\,\,\,\,A'\left( x \right) > 0\\{\rm{for }}25 < x < 50 \to A'\left( x \right) < 0\end{aligned}\]

Therefore, the function will have an absolute minimum value at \[x = 25\].

And the maximum area will be as:

\[\begin{aligned}{c}A\left( {25} \right) = 50\left( {25} \right) - {\left( {25} \right)^2}\\ = 625{\rm{ }}{{\rm{m}}^2}\end{aligned}\]

Hence, the dimensions of a rectangle for the maximum area is \[x = y = 25{\rm{ meters}}\].