Optimization is one of the most practical uses of calculus. Whether you're trying to maximize profit, minimize materials, or find the most efficient shape, derivatives help you find the optimal solution. The process involves setting up a function to optimize, finding its critical points, and determining which gives the maximum or minimum value.
🎯 In this section you will learn
- A step-by-step strategy for solving optimization problems
- How to translate word problems into mathematical functions
- How to find absolute maximum and minimum values
- Real-world examples: area, volume, cost, revenue
📌 Step-by-Step Strategy
①
Read and understand
Identify what is being optimized (maximized or minimized).
②
Draw a diagram
If applicable, sketch the situation and label variables.
③
Write the objective function
Express the quantity to be optimized as a function of one variable.
④
Find constraints
Use the given information to relate variables (if multiple).
⑤
Find critical points
Take the derivative and set it equal to zero.
⑥
Check endpoints
If the domain is a closed interval, evaluate at endpoints.
⑦
Interpret the result
State the answer in the context of the problem.
You have 100 meters of fencing to enclose a rectangular field. What dimensions maximize the area?
①
Define variables
Let $x$ = width, $y$ = length. Perimeter: $2x + 2y = 100$ ⇒ $x + y = 50$
②
Objective function (Area)
$A = x \cdot y = x(50 - x) = 50x - x^2$
③
Differentiate and find critical point
$A'(x) = 50 - 2x = 0$ ⇒ $x = 25$
④
Find $y$ and verify
$y = 50 - 25 = 25$. Second derivative: $A''(x) = -2 < 0$ ⇒ maximum.
⑤
Answer
The rectangle is a $25 \times 25$ square. Maximum area = $625$ m².
A cylindrical can holds 1000 cm³ of liquid. Find the dimensions that minimize the surface area (top and bottom included).
①
Volume constraint
$V = \pi r^2 h = 1000$ ⇒ $h = \frac{1000}{\pi r^2}$
②
Objective (Surface Area)
$S = 2\pi r^2 + 2\pi r h = 2\pi r^2 + 2\pi r \cdot \frac{1000}{\pi r^2} = 2\pi r^2 + \frac{2000}{r}$
③
Differentiate
$S'(r) = 4\pi r - \frac{2000}{r^2}$
④
Set $S'(r)=0$
$4\pi r = \frac{2000}{r^2}$ ⇒ $4\pi r^3 = 2000$ ⇒ $r^3 = \frac{500}{\pi}$ ⇒ $r = \sqrt[3]{\frac{500}{\pi}} \approx 5.42$ cm
⑤
Find $h$
$h = \frac{1000}{\pi r^2} \approx 10.83$ cm
A company sells $x$ units of a product at a price $p = 100 - 0.5x$ dollars per unit. The cost is $C(x) = 500 + 20x$. Find the quantity that maximizes profit.
①
Revenue
$R(x) = x \cdot p = x(100 - 0.5x) = 100x - 0.5x^2$
②
Profit
$P(x) = R(x) - C(x) = (100x - 0.5x^2) - (500 + 20x) = -0.5x^2 + 80x - 500$
③
Differentiate
$P'(x) = -x + 80 = 0$ ⇒ $x = 80$
④
Verify maximum
$P''(x) = -1 < 0$ ⇒ maximum at $x=80$.
⑤
Answer
Sell 80 units for maximum profit. Price = $100 - 0.5(80) = \$60$ per unit.
Find the point on the parabola $y = x^2$ that is closest to the point $(0, 3)$.
①
Distance formula
$d = \sqrt{(x-0)^2 + (x^2 - 3)^2} = \sqrt{x^2 + (x^2-3)^2}$
②
Minimize $d^2$ (easier)
$f(x) = x^2 + (x^2-3)^2 = x^2 + x^4 - 6x^2 + 9 = x^4 - 5x^2 + 9$
③
Differentiate
$f'(x) = 4x^3 - 10x = 2x(2x^2 - 5) = 0$ ⇒ $x=0$, $x = \pm\sqrt{\frac{5}{2}}$
④
Evaluate
$f(0)=9$, $f(\pm\sqrt{2.5}) \approx 6.25 + 0.25 = 6.5$ (smaller). Minimum at $x = \pm\sqrt{2.5}$.
⑤
Points
$(\sqrt{2.5}, 2.5)$ and $(-\sqrt{2.5}, 2.5)$, distance $\approx \sqrt{6.5} \approx 2.55$.
⚠️ Common Mistakes
- Using the wrong objective function — make sure you're optimizing what the problem asks.
- Forgetting constraints — if you have two variables, you need a constraint to eliminate one.
- Forgetting to check endpoints — sometimes the maximum/minimum occurs at a boundary.
- Not verifying the type of extremum — use the first or second derivative test to confirm max/min.
🔍 Key Takeaways
- Optimization = finding the best value (max or min) of a function.
- Follow the 7-step strategy: read → diagram → objective → constraint → derivative → critical points → interpret.
- Use constraints to reduce the objective function to a single variable.
- Always check domain endpoints when applicable.
- Real-world applications include area, volume, cost, profit, and distance.
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