If three sides of a rectangular park have a total length $400 \mathrm{ft}$, then the area of the park is maximum when the length (in ft ) of its longer side is

When we say "three sides of a rectangular park have a total length of 400 ft," we need to visualize this carefully.

A rectangle has 4 sides: 2 lengths and 2 widths. If we're told about 3 sides totaling 400 ft, this means two sides of one dimension + one side of the other dimension = 400 ft.

Let's call the dimensions:

Length = $l$ ft

Width = $b$ ft

So we have: $2l + b = 400$

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Area of rectangle = length × width = $l \times b$

From our constraint $2l + b = 400$, we can express $b$ in terms of $l$:

$b = 400 - 2l$

Substituting this into our area formula:

$\text{Area} = l \times b = l(400 - 2l) = 400l - 2l^2$

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Here's where we use a clever mathematical trick. Let's rewrite our area expression:

$\text{Area} = l(400 - 2l) = l \times 2(200 - l) = 2l(200 - l)$

Key Insight: For any two positive numbers that have a fixed sum, their product is maximum when they are equal.

In our expression $2l(200 - l)$, we can think of this as $2 \times [l \times (200 - l)]$.

The product $l \times (200 - l)$ will be maximum when:

$l = 200 - l$

$2l = 200$

$l = 100$

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Now we can find the width:

$b = 400 - 2l = 400 - 2(100) = 400 - 200 = 200$

Important Check: Since $b = 200$ ft and $l = 100$ ft, the longer side is actually the width!

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This approach uses the mathematical principle that for a fixed perimeter, the product of two terms is maximized when they are as close to equal as possible.

We rewrote our area as $2l(200-l)$ and maximized $l(200-l)$ by setting $l = 200-l$.

Therefore, the length of the longer side is 200 feet.