Welcome back to Professor Baker's Math Class! In Section 4-7, we are tackling one of the most useful and applicable topics in Calculus: Optimization Problems. Whether you are an engineer trying to minimize the cost of materials, a business owner maximizing revenue, or a farmer planning a field, these methods allow us to find the "best" possible values for a given situation.
The 6-Step Strategy for Success
As outlined in the lecture notes, word problems can be intimidating, but they become manageable when you follow a structured approach. Here is the six-step method we discussed in class:
- Understand the Problem: Read carefully. Identify what is unknown, what is given, and exactly what needs to be maximized or minimized.
- Draw a Diagram: Visualizing the problem is crucial. Label your diagram with the given quantities.
- Introduce Notation: Assign symbols to your variables. For example, let $Q$ be the quantity to optimize, $A$ for area, or $C$ for cost.
- Express the Function: Write an equation for the quantity $Q$ in terms of your other symbols.
- Reduce to One Variable: Use the constraints (given information) to rewrite your equation so that $Q$ is a function of only one variable (e.g., $Q = f(x)$). Don't forget to determine the domain!
- Apply Calculus: Use the methods from Sections 4.1 and 4.3 (finding critical numbers where $f'(x)=0$) to find the absolute maximum or minimum.
Class Example 1: Maximizing Area
We started with a classic geometry problem: A farmer has 2400 ft of fencing to enclose a rectangular field bordering a river. No fence is needed along the river. What dimensions yield the largest area?
Here, we identified two equations:
- Constraint (Perimeter): $2400 = L + 2W$ (since one side is the river).
- Optimization (Area): $A = LW$.
By solving the constraint for $L$, we get $L = 2400 - 2W$. Substituting this into the area equation gives us a function of a single variable:
$$A(W) = (2400 - 2W)W = 2400W - 2W^2$$Taking the derivative $A'(W) = 2400 - 4W$ and setting it to zero gave us a width of $W=600$. This confirms that optimization allows us to find the precise dimensions for maximum efficiency.
Key Tip: Minimizing Distance
In the example regarding the point on the parabola $y^2 = 2x$ closest to $(1, 4)$, we used a handy trick. The distance formula involves a square root: $d = \sqrt{(x-x_1)^2 + (y-y_1)^2}$.
Pro Tip: To minimize distance $d$, you can simply minimize the square of the distance, $d^2$. This removes the messy square root from your derivative process but results in the same critical $x$ or $y$ value. As shown in the notes, we optimized:
$$f(y) = \left(\frac{1}{2}y^2 - 1\right)^2 + (y-4)^2$$Business & Physics Applications
We also explored how these concepts apply to economics and electricity:
- Maximizing Revenue: In the TV monitor example, we derived a Revenue Function $R(x)$ based on price changes. Remember, Revenue = (Price per unit) $\times$ (Number of units sold).
- Electrical Power: We examined a resistor problem where Power $P = \frac{E^2 R}{(R+r)^2}$. By finding $P'(R)$, we discovered that power is maximized when the external resistance equals the internal resistance ($R = r$).
Be sure to review the attached PDF for the full step-by-step derivation of the Cylindrical Can problem (Example 2) and the complete algebraic breakdown of the Power output problem. Keep practicing those derivatives, and remember to check your endpoints if you are working on a closed interval!