Welcome back to Professor Baker’s Math Class! In Section 4.8, we arrive at one of the most practical and satisfying applications of Calculus 1: Optimization Problems. Whether you are an engineer trying to minimize the material used for a can, or a business owner trying to maximize revenue, the derivative is your best friend.
Optimization is all about finding the absolute maximum or minimum values of a function. However, unlike previous sections where the function was given to you, here you often have to build the function yourself from a word problem.
The 6-Step Strategy for Success
As outlined in the class notes, having a structured approach is crucial. Here is the roadmap to solving these problems:
- Understand the Problem: Read carefully. Identify what is being maximized or minimized (the Objective) and what limitations exist (the Constraints).
- Draw a Diagram: Visualizing the problem is half the battle.
- Introduce Notation: Assign variables to the quantities ($x$, $y$, $r$, $h$, etc.).
- Express the Quantity ($Q$): Write an equation for the value you want to optimize in terms of your variables.
- Reduce to One Variable: Use the given constraints to eliminate extra variables. You want your objective function, $f(x)$, to have only one input variable.
- Use Calculus: Find the absolute maximum or minimum using derivatives (finding critical numbers where $f'(x) = 0$).
Example 1: The Farmer and the River
A classic problem involves a farmer with 2400 ft of fencing who wants to fence off a rectangular field bordering a straight river. No fence is needed along the river. What dimensions yield the largest area?
The Constraint: The total fencing available is 2400 ft. If $y$ is the width (perpendicular to the river) and $x$ is the length, our constraint is:
$$2y + x = 2400$$
The Objective: We want to maximize Area ($A$).
$$A = xy$$
By solving the constraint for $x$ ($x = 2400 - 2y$) and substituting it into the area equation, we get a function of one variable:
$$A(y) = (2400 - 2y)y = 2400y - 2y^2$$
Taking the derivative, $A'(y) = 2400 - 4y$, and setting it to zero gives us $y = 600$. Plugging this back in, we find $x = 1200$. Thus, the dimensions for the largest area are $600 \text{ ft} \times 1200 \text{ ft}$.
Example 2: Minimizing Manufacturing Costs
We also looked at 3D geometry. Suppose we need to make a cylindrical can that holds $1 \text{ L}$ ($1000 \text{ cm}^3$) of oil. We want to minimize the cost of metal, which means minimizing the Surface Area (SA).
- Constraint (Volume): $\pi r^2 h = 1000$. This allows us to solve for height: $h = \frac{1000}{\pi r^2}$.
- Objective (Surface Area): $SA = 2\pi r^2 + 2\pi r h$.
By substituting $h$ into the SA equation, we get a function entirely in terms of the radius, $r$. We can then take the derivative, set it to zero, and solve for the optimal radius to keep costs low.
Example 3: Business Application
Calculus isn't just for geometry; it maximizes profit too. We examined a store selling monitors. We found that for every rebate offered, sales volume increased. By setting up a Revenue function:
$$R(x) = (\text{Price}) (\text{Quantity})$$ $$R(x) = (350 - 10x)(200 + 20x)$$
We expanded this, found the derivative $R'(x)$, and determined the optimal rebate amount ($x$) to generate the highest possible revenue.
Key Takeaway: Don't let the word problems intimidate you. Draw your picture, identify your constraint, and let the calculus do the heavy lifting!