Park Designer Problem

Question

A park designer has 300 meters of fencing and wants to create a rectangular play area. One of the longer sides of the rectangle will be adjacent to an existing wall, so only the remaining three sides (two short sides and the opposite long side) need to be fenced.

(a) What dimensions will maximize the area of the play space?

(b) If the designer wants to divide the area into two equal parts by adding a fence parallel to the two short sides, what are the new optimal dimensions for the maximum area?

By: <a href='https://basketrandom.org'>basket random</a>

Answer 1

I hope you agree the following expression is fair for what we are being asked:

300 = 2x + y

2 short walls called x plus one big wall called y will add up to 300 units.

If you agree with this, then let me also say that:

y = 300 - 2x

Let us think about a different expression. We are optimizing for area in this problem. Optimization means we are finding the absolute maximum or minimum of a function. We would be making good progress if we are able to find a function for the area of this fence enclosure.

We can model the area of a rectangle by multiplying one short side by one long side. Since one of the short sides of the fence is x and the long side is y, we can say that:

The area of the fence = x*y

This is a good model, except it has none of the restrictions that our other expression has.

We want a model of area where x*y, but we also want one where y = 300 - 2x, which is the rule for how big x is in comparison to y.

We can add this rule by replacing y with (300 - 2x). So:

The area of the fence = (300 - 2x)*x

= 300x - 2x²

Now this will find all values of x that satisfy all our rules. I am going to call this function f(x).

f(x) = 300x - 2x²

We must find the maximum value of this function. We can do this by finding when the slope of the function is 0, which works since this is a parabola, and there is only one point on a parabola that will have a slope of 0 (when the function is not increasing or decreasing).

We can do this by finding the zeros of the derivative of f(x), as the derivative f'(x) of a function f(x) is just a function that will tell you the slope at any given point in f(x). Using the power rule, The derivative of f(x) is:

300 - 4x

This equation evaluates to 0 at x = 75. This means that when you plug 75 into f'(x), it comes out as 0 slope. That means when you plug 75 into f(x), it will be at a point where the slope is zero, or in other words, the maximum, where the function is no longer increasing or decreasing.

So our original function is:

f(x) = 300x - 2x²

I will plug in 75:

f(75) = 300(75) - 2(75)²

f(75) = 22500 - 11250

f(75) = 11250

This is as big as our fence is going to get. 11,250 meters²!

This area can only be attained when x = 75. Remember that x is one of the two shorter sides in this equation where all three sides at up to 300 meters:

300 = 2x + y

So:

300 = 2(75) + y

300 = 150 + y

150 = y

So the two short sides are 75 meters long and the long side is 150 meters long for a total area of 11,250 meters².

For part B, you are following the same logic as above except now you have 3 short sides, so your new expression would be:

300 = 3x + y

And plugging this into:

Area of fence = xy

Yields:

Area of fence = x*(300 - 3x)

Area of fence = 300x - 3x²

This is our new function I will call g(x):

g(x) = 300x - 3x²

The derivative of this is:

g'(x) = 300 - 6x

Which evaluates to 0 at x = 50

So our function reaches its maximum at x = 50.

In our original function, g(50) evaluates to:

g(50) = 300(50) - 3(50)²

g(50) = 15000 - 7500

g(50) = 7500

Or 7,500 meters²

So this is our greatest possible area with the dividing fence. Since x = 50 here, and that is one of the three small sides, we know the large side will still be 150 meters long.

So the three short sides are 50 meters long and the long side is 150 meters long for a total area of 7,500 meters².