Optimization Word Problems: Finding the Equation and Domain

To solve optimization problems, follow the steps below.

If you are presented with a word problem, turns the situation into a math problem. Consider the optimization issue that has been presented and ask yourself the question $-$ what function am I optimizing? The optimization can be a maximization or a minimization.

Next, once you have your function $f(x)$ to find the maxima/minima of, realize that you can only do this if you find the derivative of $f(x)$. We may need higher-order derivatives but start by finding $f'(x)$.

Once you have found all $x$ for which $f'(x)=0$, you know where there are local minima and maxima. That might me all you are interested in. But if not, you want to read on.

Don't forget to check if the domain you are considering contains the values of $x$ you obtained. If none of the values you found are in the domain, there is no point of interest. If, however, there is, you know there is at least one absolute extremum.

One question we might ask, for each value of $x$, is $-$ at this point, do we encounter a local minimum or maximum? You should know the answer to this!

The Second Derivative Test! If you find that at a point $\left( x_0, f(x_0) \right)$, we have $f'(x_0)=0$ and $f''(x_0)<0$, we have a local maximum. If we have $f'(x_0)=0$ and $f''(x_0)>0$, we have a local minimum. The intuition for this is explained in the Second Derivative Test section.

So what's left? Of course, the equality case. If $f'(x_0) = 0$ and $f''(x_0) = 0$, the test is not informative, and instead you should look for nearby values of $f(x)$ and try to decide if, at $x=x_0$, you have a local maximum or a local minimum.

Now that we know all the basics, let's try our hands at an optimization problem. First, you solve it. Then, we discuss it. The below problem is a classic so if we were to give credits to a particular source, it would be unfair.

Solution. First, let's ignore units, since we are working in feet and we know it. Let the horizontal dimension be $x$ and let the vertical dimension be $y$. Let us say the side missing is a horizontal side. Thus, the perimeter is $x + 2y$ and the area is $xy$. We know that the perimeter is $500$, so we have the equation $$x + 2y = 500 \implies x = 500 - 2y.$$Let the area function, in terms of $y$, be $A(y)$. Then, we have $$A(y) = xy = \left( 500 - 2y \right) \cdot y = 500y - 2y^2.$$Now, if you have taken algebra, you realize that this is a parabola and you can probably optimize $A(y)$ without calculus, but this is how we practice!

First, we note that $A'(y) = 500 - 4y$ and we set $A'(y) = 0$ and solve to get $$500 - 4y = 0 \implies y = 125 \implies x = 250.$$Hence, the area is optimized when $(x,y)=(250,125)$. If we wanted to verify that this is indeed a maximum, we could use the second derivative test to get $$A''(y) = -4 < 0, \qquad \forall y.$$Hence, we indeed have a maximum.

The maximum area is, thus, $250 \cdot 125 = \boxed {31250}$ (square feet).