Maximizing area of rectangular portion of athletic field

[CinnamonStix]

An athletic field is to be built in the shape of a rectangle x m long capped by semicircular regions of radius r m at the two ends. The field is to be bounded by a 200 m racetrack.

Express the area of the rectangular portion of the field as a function of x alone.

What value of x gives the rectangular portion the largest possible area? Thanks for the help :)

 

[MarkFL]

Re: Calculus help

Hello, and welcome to MHB, CinnamonStix!

We like for our users to show what they've done so we know where you are stuck. But, let's assume you are asking how to begin. All measures will be in terms of meters.

We know the area $A$ of the rectangular portion of the field in terms of $x$ and $r$:

\(\displaystyle A(r,x)=2rx\)

We also know the perimeter of the entire field in terms of $x$ and $r$:

\(\displaystyle P(r,x)=2x+2\pi r=2(x+\pi r)\)

We are given an additional piece of information, and that is the perimeter of the entire field is to be 200 m. And so we may state:

\(\displaystyle 2(x+\pi r)=200\)

If you solve the above equation for $r$, and then substitute into the area function, you will have the area in terms of $x$ alone...what do you get?

 

[CinnamonStix]

So we would have..
2(x+pi*r)=200
(x+pi*r)=100
r=(100-x)/pi

and after plugging it into A=2rx we would have the answer of...
A= 2x((100-x)/pi)

For part 2,

I believe we take the first derivative of A= 2x((100-x)/pi), set it to 0 and solve for x.
We would end up with x=50m.
I just don't understand why we want to find the derivative in the first place. Can somebody explain please

 

[MarkFL]

So we would have..
2(x+pi*r)=200
(x+pi*r)=100
r=(100-x)/pi

and after plugging it into A=2rx we would have the answer of...
A= 2x((100-x)/pi) Can you also help me out with part 2? I don't even know how to begin

Yes, we have:

\(\displaystyle A(x)=2\left(\frac{100-x}{\pi}\right)x=\frac{2}{\pi}x(100-x)\)

Before we use the calculus to optimize the area function, let's take a moment to look at the area function. We see that it is quadratic, and that it opens downward. Thus, the vertex will be at the maximum, and additionally, we know the roots. For a parabolic function, the vertex lies along the axis of symmetry, and where does the axis of symmetry reside in relation to the roots?

 

[MarkFL]

To follow up, the axis of symmetry will lie midway between the two roots, and in our case the roots are:

\(\displaystyle x\in\{0,100\}\)

Thus, the axis of symmetry is the line:

\(\displaystyle x=\frac{0+100}{2}=50\)

And so, this is the value of $x$ that will maximize the area of the rectangular portion of the field. To demonstrate this using the calculus, we may compute:

\(\displaystyle A'(x)=\frac{2}{\pi}\left((100-x)+(-x)\right)=\frac{4}{\pi}(50-x)=0\implies x=50\)

Now, observing that:

\(\displaystyle A''(x)=-\frac{4}{\pi}<0\)

We may conclude (from the second derivative test) that the critical value is at the global maximum.