Sometimes we have a function and we just really want to know what its high point or low point is in terms of [latex]y[/latex]-value. The high point is called a maximum, and the low point is called a minimum. For most functions, these points occur when the derivative is zero or undefined (we talked about why this is briefly in a previous section).
We will follow the maxim “optimization happens when the derivative is zero”. First we find the derivative using the power rule [latex]h'(t) = -10t + 20[/latex]. Then we set this equal to zero, so we solve
\begin{align*}
-10t + 20 & = 0 \\
-10t & = – 20 \\
t & = \boxed{2}
\end{align*}
Hence, the height is maximized when the time is equal to [latex]2[/latex] seconds. At this point, the height of the ball is [latex]h(2) = -5(2)^2 + 20(2) + 10 = -20 + 40 + 10 = 30[/latex] meters. Here’s a rough sketch based on what we know about this function:
Let’s do anther optimization example:
[latex]C(x) = \frac{500}{x} + 0.001x.[/latex]
At what value [latex]x[/latex] is the cost per item minimized? What is the cost at this value of [latex]x[/latex]?
To solve this problem, we find [latex]C'(x)[/latex]:
[latex]C'(x) = \frac{d}{dx} 500 x^{-1} + 0.001 x = -500x^{-2} + 0.001[/latex]
We then set this equal to zero and solve:
\begin{align*}
-500x^{-2} + 0.001 & = 0 \\
-500x^{-2} & = -0.001 \\
x^{-2} & = 0.000002 \\
1 & = 0.000002 x^2 \\
500,000 & = x^2 \\
\sqrt{500,000} & = x \approx \boxed{707.10}
\end{align*}
So we see about [latex]707[/latex] action figures is the best number to choose. The cost at this point (by plugging [latex]x = 707[/latex] into the original equation) is just [latex]\boxed{\$1.41}[/latex]. Not too bad!