# If a rectangular area is required to have a perimeter of "100 m", what dimensions maximize the area?

Aug 18, 2016

To do this, you have to assume a perimeter of $\text{100 m}$, and utilize that information to generate the largest possible area.

This can algebraically be written as:

$\boldsymbol{A = l \times h}$
(area as related to length and height)

$\boldsymbol{P = 2 \times l + 2 \times h = 100}$
(perimeter as related to length and height)

When you solve for $l$ in the perimeter equation, you should get:

$2 l = 100 - 2 h$

$\textcolor{g r e e n}{l = 50 - h}$

If you think about the factors that multiply to give you a perimeter of $100$, pick some easy ones, and you can have:

$2 \times 5 + 2 \times 45 = 100$
$2 \times 10 + 2 \times 40 = 100$
$2 \times 15 + 2 \times 35 = 100$
$2 \times 20 + 2 \times 30 = 100$
$2 \times 25 + 2 \times 25 = 100$

Notice how if $h = 45$, then $l = 50 - h = 5$, and so on. Just pick multiple values of the height $h$, and use the corresponding value of the length $l$ to look at dimension combinations.

And if you then use these lengths and heights to calculate the area you'd get:

$5 \times 45 = 225$
$10 \times 40 = 400$
$15 \times 35 = 525$
$20 \times 30 = 600$
$25 \times 25 = \textcolor{b l u e}{625}$

If you go any further, you would see that the length/height combinations have been exhausted and you would only have other symmetrical combinations (e.g. $5 \times 45$ vs. $45 \times 5$).

This means the largest rectangular field in area has dimensions of $\textcolor{b l u e}{\text{25 m}}$ $\times$ $\textcolor{b l u e}{\text{25 m}}$.