# Scott's question at Yahoo Answers regarding optimization with constraint

• MHB
• MarkFL
In summary: Therefore, in summary, we can determine that the largest possible area for the athletic field is achieved when the rectangle is 100 meters long and the semicircular regions have a radius of 100/pi meters.
MarkFL
Gold Member
MHB
Here is the question:

Scott said:
An athletic field is to be built in the shape of a rectangle x units long capped by semicircular regions of radius r at the two ends?

The field is to be bounded by a 400 meter running track. What values of x and r will give the rectangle the largest possible area?

I have posted a link there to this thread so the OP can view my work.

Hello Scott:

The objective function (that which we wish to optimize) is the area of the rectangle, and is as follows:

$$\displaystyle f(r,x)=2rx$$

And if we let $P$ be the perimeter of the field, we may state the constraint:

$$\displaystyle g(r,x)=2x+2\pi r-P=0$$

Now, we have two ways to proceed. The first is to solve the constraint for one of the two variables $r$ or $x$, and then substitute into the objective function so that we have a function in one variable and then optimize by differentiation. Sol if we solve the constraint for $r$, we obtain:

$$\displaystyle r=\frac{P-2x}{2\pi}$$

And the substituting into the objective function, we obtain:

$$\displaystyle f(x)=2\left(\frac{P-2x}{2\pi}\right)x=\frac{x(P-2x)}{\pi}$$

Now, at this point we see we have a quadratic function opening downwards and we could use a precalculus technique to find the maximum value. We see the roots are:

$$\displaystyle x=0,\,\frac{P}{2}$$

We know then that the axis of symmetry lies midway between these roots, and so the axis of symmetry is at :

$$\displaystyle x=\frac{P}{4}$$

And so we have:

$$\displaystyle r=\frac{P-2\left(\frac{P}{4}\right)}{2\pi}=\frac{P}{4\pi}$$

Now, let's try differentiation. Recall:

$$\displaystyle f(x)=\frac{x(P-2x)}{\pi}=\frac{1}{\pi}\left(Px-2x^2\right)$$

Hence, differentiating with respect to $x$, and equating the result to zero to obtain the critical value(s), we obtain:

$$\displaystyle f'(x)=\frac{1}{\pi}\left(P-4x\right)=0\implies x=\frac{P}{4}$$

And like before, we find:

$$\displaystyle r=\frac{P}{4\pi}$$

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

Since the objective function is concave down everywhere, we know the critical value is at the global maximum.

Now, let's examine a multivariable method: Lagrange multipliers. Recall, we have:

The objective function:

$$\displaystyle f(r,x)=2rx$$

Subject to the constraint:

$$\displaystyle g(r,x)=2x+2\pi r-P=0$$

Hence, we obtain the system:

$$\displaystyle 2x=\lambda(2\pi)$$

$$\displaystyle 2r=\lambda(2)$$

Hence, this implies:

$$\displaystyle \lambda=\frac{x}{\pi}=r$$

Substituting for $r$ into the constraint, we obtain:

$$\displaystyle 2x+2\pi\left(\frac{x}{\pi}\right)-P=0$$

$$\displaystyle 4x=P$$

$$\displaystyle x=\frac{P}{4}\implies r=\frac{P}{4\pi}$$

So, given that the objective function is zero for $x=0$, we may conclude that this critical point is at a maximum.

Thus, we have shown in various ways that:

$$\displaystyle f_{\max}=f\left(\frac{P}{4},\frac{P}{4\pi}\right)$$

Using the given value $P=400\text{ m}$, we then may conclude that the rectangular portion of the field is maximized for:

$$\displaystyle x=\frac{400\text{ m}}{4}=100\text{ m}$$

$$\displaystyle r=\frac{400\text{ m}}{4\pi}=\frac{100}{\pi}\text{ m}$$

This means that the straight portions of the track are equal in length to the curved portions.

## 1. What is optimization with constraint?

Optimization with constraint is a mathematical process used to find the best possible solution to a problem while adhering to certain limitations or restrictions. It involves maximizing or minimizing a function while taking into account constraints that must be satisfied.

## 2. How is optimization with constraint used in real life?

Optimization with constraint has a wide range of applications in fields such as engineering, economics, and operations research. It can be used to optimize the design of structures, maximize profits in business, and improve efficiency in transportation and logistics, among other things.

## 3. What are some common constraints in optimization problems?

Constraints can take many forms and vary depending on the specific problem being solved. Some common constraints include budget limitations, time constraints, physical limitations, and resource availability.

## 4. What are the different methods used for optimization with constraint?

There are several methods used for optimization with constraint, including linear programming, non-linear programming, and dynamic programming. Each method has its own advantages and is suited for different types of problems.

## 5. How is optimization with constraint different from traditional optimization?

The main difference between optimization with constraint and traditional optimization is the presence of restrictions or limitations in the former. In traditional optimization, the goal is simply to find the best solution, whereas in optimization with constraint, the solution must also satisfy certain constraints.

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