Optimization in Calculus centers around Word Problems that require looking for either the largest or smallest possible value(s) that a function can use. For example, a company might want to maximize revenue or minimize travel time. This lesson is, to a degree, a continuation of the concepts that were covered in Extreme Values.
When working with optimization, there's normally a constraint that needs to be taken into account. This constraint is a condition (ie a fixed value like a measurement) that must be true regardless of the solution. If there are multiple solutions that we can determine, this constraint helps narrow the actual solutions to just 1.
A \(400\;[\text{m}]\) track is to be constructed of 2 straight-ways and 2 semicircular ends. The straight-ways can be no less than \(100\;[\text{m}]\) long. What radius would produce the maximum area?
We need to determine a radius, \(r\), that can provide the maximum Area in a track with a Perimeter of \(400\;[\text{m}]\).
First, we can sketch a diagram to help represent this problem visually:
We can also represent the total Area of the track algebraically as:
Next, we can use the respective Perimeters to solve for \(L\):
\(P = 2L + 2\pi r\)
\(400 = 2L + 2\pi r\)
\(2L = 400 - 2\pi r\)
\(\cfrac{2L = 400 - 2\pi r}{2}\)
\(L = 200 - \pi r\)
Then, we can substitute this value for \(L\) into the original Area formula:
\(A = 2r(200 - \pi r) + \pi r^2\)
\(A = 400r - 2\pi r^2 + \pi r^2\)
\(A = 400r - \pi r^2\)
After, we can determine the respective domains for length and radius:
\(L \geq 100\)
\(L∈(100, +\infty)\)
\(200 - \pi r \geq 100\)
\(200 - 100 \geq \pi r\)
\(\cfrac{100}{\pi} \geq \cfrac{\pi r}{\pi}\)
\(r \leq \cfrac{100}{\pi}\)
\(r∈ \left(0, \cfrac{100}{\pi}\right)\)
Now, in order to determine the critical points, we can find the derivative of the Area function. Then, we can set the function equal to \(0\) and solve for the critical points:
\(A' = 400 - 2\pi r\)
\(0 = 400 - 2\pi r\)
\(2\pi r = 400\)
\(r = \cfrac{200}{\pi}\)
We can't use the value we calculated above since it doesn't fit into the domain. We will instead calculate the absolute max by substituting the endpoint values \((0\) and \(100)\) into the original function:
\(A(0) = 400(0) - \pi (0)^2\)
\(A(0) = 0 \;[\text{m}^2]\)
\(A \left(\cfrac{100}{\pi}\right) = 400\left(\cfrac{100}{\pi}\right) - \pi \left(\cfrac{100}{\pi}\right)^2\)
\(A \left(\cfrac{100}{\pi}\right) = 9554.1\;[\text{m}^2]\)
Since we have determined that \(r\) is roughly \(31.83\;[\text{m}]\), we can substitute this value into the Length equation to verify that it's at least \(100\;[\text{m}]\):
\(L = 200 - \pi (31.83)\)
\(L = 200 - 100\)
\(L = 100\;[\text{m}]\)
Therefore, we can determine that a radius of \(31.83\;[\text{m}]\) is needed to produce the maximum area.
A cardboard box with a square base is to have a Volume of \(8\;[\text{L}]\), (\(1\;[\text{L}] = 1000\;[\text{cm}^3]\)).