Optimization
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.
Steps for Solving Optimization Problems
- Identify what needs to maximized or minimized and what the constraints are
- Use diagrams (if necessary) to help visualize the problem
- Determine the variables and formula that is to be optimized
- Narrow the # of independent variables to 1 that the formula can be expressed in the form of \(f(x)\)
- Find the respective Domains (in terms of independent variables)
- Find Critical Points in Domain. Determine the function's derivative and set it equal to 0
- Show it is either a Max or Min. Use either the First or Second Derivative Tests and compare the critical points to the intervals to determine which is the solution
- Answer the question and provide proper units to the solution
Example
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:
\(A = 2rL + \pi r^2\)
We can start with finding the Length, \(L\), to help determine one of the unknown values.
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 \(\boldsymbol{31.83\;[\textbf{m}]}\) is needed to produce the maximum area.
A cardboard box with a square base is to have a Volume of \(8\;[\text{L}]\), (\(1\;[\text{L}] = 1000\;[\text{cm}^3]\)).
- Find the dimensions of the box that will minimize the amount of cardboard use.
- The cardboard for the box costs \(0.1¢/\text{cm}^2\) but the cardboard for the top and bottom is thicker so it costs 3 times as much. Find the dimensions of the box that will minimize the cost.
i. We need to determine the base, \(b\), and height, \(h\), of a Square Prism with a volume of \(8000\;[\text{cm}^3]\).
First, we can sketch a diagram to help represent this problem visually:
We can represent the area algebraically as:
\(A = 2b^2 + 4bh\)
We can start with finding the height, \(h\) to help determine one of the unknown values.
Next, we can use the formula for the volume of a Square Prism to solve for \(h\):
\(V = b^2h\)
\(8000 = b^2h\)
\(h = \cfrac{8000}{b^2}\)
We can substititue \(\cfrac{8000}{b^2}\) for \(h\) into the original equation and expand to determine our new equation:
\(A = 2b^2 + 4bh\)
\(A = 2b^2 + 4b \left(\cfrac{8000}{b^2}\right)\)
\(A = 2b^2 + \cfrac{32000}{b}\)
Then, we can determine the respective domains for the base and height:
\(h > 0\)
\(h∈(0,+∞)\)
\(\cfrac{8000}{b^2} > 0\)
\(b > 0\)
\(b∈(0,+∞)\)
After, we can determine the critical points. We can do this by determining the derivative of the original function:
\(A = 2b^2 + 32000b^{-1}\)
\(A' = 4b - 32000b^{-2}\)
\(A' = 4b - \cfrac{32000}{b^2}\)
We can now set the function equal to \(0\). We can also give both terms the same denominator and fully factor the expression to find the critical points:
\(0 = \cfrac{4b}{1} - \cfrac{32000}{b^2}\)
\(0 = \cfrac{(4b)(b^2) - 32000}{b^2}\)
\(0 = \cfrac{4b^3 - 32000}{b^2}\)
\(0 = \cfrac{4(b^3 - 8000)}{b^2}\)
\(0 = \cfrac{4(b-20)(b^2+20b+400)}{b^2}\)
We can now solve for the critical points:
\(b - 20 = 0\)
\(b = 20 \;[\text{cm}]\)
\(b = 0\)
Therefore, we can determine that our critical point lies at \(b = 20\). We can't use \(b=0\) since it doesn't fit in the domain.
In order to show that \(b = 20\) is a min point, we can plug this value into the original function along with 2 close values, one larger and one smaller:
\(A = 2b^2 + 32000b^{-1}\)
\(A(19) = 2(19)^2 + 32000(19)^{-1}\)
\(A(19) = 2406.21\)
\(A(20) = 2(20)^2 + 32000(20)^{-1}\)
\(\textcolor{red}{A(20) = 2400}\)
\(A(21) = 2(21)^2 + 32000(21)^{-1}\)
\(A(21) = 2405.81\)
Since we have determined that \(b=20\;[\text{cm}]\), we can substitute this value into the height equation to determine \(h\):
\(h = \cfrac{8000}{(20)^2}\)
\(h = \cfrac{8000}{400}\)
\(h = 20\;[\text{cm}]\)
Therefore, we can determine that \(\boldsymbol{b = 20\;[\textbf{cm}]}\) and \(\boldsymbol{h = 20\;[\textbf{cm}]}\).
ii. In order to determine how to minimize the cost, we must first determine the equation for the original function:
\(C = 0.3(2b^2) + 0.1(4bh)\)
\(C = 0.6b^2 + 0.4b \left(\cfrac{8000}{b^2}\right)\)
\(C = 0.6b^2 + \cfrac{3200}{b}\)
Next, we can determine the domain of the base and height:
\(b∈(0,+∞)\)
\(h∈(0,+∞)\)
Then, we can determine the critical points by finding the derivative of the original function:
\(C= 0.6b^2 + 3200b^{-1}\)
\(C'= 1.2b - 3200b^{-2}\)
\(C'= 1.2b - \cfrac{3200}{b^2}\)
We can now set this function equal to \(0\). After, we can set a common denominator to the whole expression and fully factor it to find the critical points:
\(0 = \cfrac{1.2b}{1} - \cfrac{3200}{b^2}\)
\(0 = \cfrac{(1.2b)(b^2) - 3200}{b^2}\)
\(0 = \cfrac{1.2b^3 - 3200}{b^2}\)
\(0 = \cfrac{(b-13.87)(b^2+13.87b+192.3)}{b^2}\)
We can now determine the critical points:
\(b - 13.87 = 0\)
\(b = 13.87 \; [\text{cm}]\)
\(b = 0\)
Therefore, we can determine that \(b = 13.87\). We can't use \(b=0\) since it doesn't fit within the domain.
In order to show that this is the Min point, we can plug this value into the original function along with 2 close values, one larger and one smaller.
\(C= 0.6b^2 + 3200b^{-1}\)
\(C(12) = 0.6(12)^2 + 3200(12){-1}\)
\(C(12) = 353.07\)
\(C(13.87) = 0.6(13.87)^2 + 3200(13.87){-1}\)
\(\textcolor{red}{C(13.87) = 346.14}\)
\(C(15) = 0.6(15)^2 + 3200(15){-1}\)
\(C(15) = 348.33\)
Since we have determined that \(b=13.87\), we can substitute this value into the height equation to determine \(h\):
Finally, we can substitute this value into the height equation to determine \(h\):
\(h = \cfrac{8000}{(13.87)^2}\)
\(h = \cfrac{8000}{192.3769}\)
\(h = 41.6\;[\text{cm}]\)
Therefore, we can determine that \(\boldsymbol{b = 13.87\;[\textbf{cm}]}\) and \(\boldsymbol{h = 41.6\;[\textbf{cm}]}\) in order minimize the cost.