The Logic of the “Peak”#
Optimization is the most practical part of Calculus. It involves finding the “best” possible outcome — like the maximum volume of a box or the minimum cost of production.
The Key Insight#
At a maximum or minimum, the gradient is zero ($f'(x) = 0$). This is because at the very top of a hill or the very bottom of a valley, the tangent line is perfectly horizontal.
1. The 5-Step Strategy#
Every optimization problem follows the same structure:
Step 1: Identify the variable to optimize#
Read the question carefully — what must be maximized or minimized? (Volume, area, profit, cost, distance, etc.)
Step 2: Write a formula for it#
Express the quantity to optimize as a function. Often this involves geometry formulas.
Step 3: Reduce to ONE variable#
If your formula has two variables (like $r$ and $h$), use a constraint (a second equation given in the problem) to eliminate one variable by substitution.
Step 4: Differentiate and set equal to zero#
Find $f'(x) = 0$ and solve for $x$.
Step 5: Answer the actual question#
Don’t stop at finding $x$! Substitute back into the original formula to find the maximum/minimum value. Read the last line of the question to check what’s being asked.
2. Worked Example 1: Open Box from Cardboard#
A rectangular sheet of cardboard is 20 cm by 12 cm. Equal squares of side $x$ cm are cut from each corner and the sides are folded up to make an open box. Find the value of $x$ that gives the maximum volume.
Step 1: Optimize Volume.#
Step 2: After cutting and folding:#
- Length: $20 - 2x$
- Width: $12 - 2x$
- Height: $x$
Step 3: Already in one variable ($x$). Expand:#
$V = x(240 - 40x - 24x + 4x^2) = x(4x^2 - 64x + 240)$
$V = 4x^3 - 64x^2 + 240x$
Step 4: Differentiate:#
$V' = 12x^2 - 128x + 240$
Set $V' = 0$: $12x^2 - 128x + 240 = 0$
Divide by 4: $3x^2 - 32x + 60 = 0$
Using the quadratic formula: $x = \frac{32 \pm \sqrt{1024 - 720}}{6} = \frac{32 \pm \sqrt{304}}{6} = \frac{32 \pm 17.44}{6}$
$x = 8.24$ or $x = 2.43$
Step 5: Check validity#
Since the width is 12 cm, we need $12 - 2x > 0$, so $x < 6$. Therefore $x = 8.24$ is rejected (it would give a negative width).
$x = 2.43$ cm gives the maximum volume.
$V = 2.43(20 - 4.86)(12 - 4.86) = 2.43 \times 15.14 \times 7.14 = 262.7 \text{ cm}^3$
Verify it’s a maximum: $V''(x) = 24x - 128$. At $x = 2.43$: $V'' = 58.32 - 128 = -69.68 < 0$ ✓ (concave down = maximum).
3. Worked Example 2: Cylinder with Fixed Surface Area#
A closed cylinder has a total surface area of $600\pi$ cm². Show that the volume is $V = 300\pi r - \pi r^3$, and find the radius that gives the maximum volume.
Step 2: Formulas:#
- Surface area: $SA = 2\pi r^2 + 2\pi rh = 600\pi$
- Volume: $V = \pi r^2 h$
Step 3: From the constraint, express $h$ in terms of $r$:#
$2\pi r^2 + 2\pi rh = 600\pi$
Divide by $2\pi$: $r^2 + rh = 300$
$rh = 300 - r^2$
$h = \frac{300 - r^2}{r}$
Substitute into Volume:#
$V = \pi r^2 \cdot \frac{300 - r^2}{r} = \pi r(300 - r^2) = 300\pi r - \pi r^3$ ✓
Step 4:#
$V' = 300\pi - 3\pi r^2$
Set $V' = 0$: $300\pi - 3\pi r^2 = 0$
$3\pi r^2 = 300\pi$
$r^2 = 100$
$r = 10$ cm (reject $r = -10$)
Step 5:#
$h = \frac{300 - 100}{10} = 20$ cm
$V = \pi(10)^2(20) = 2000\pi \approx 6283.2$ cm³
Verify: $V'' = -6\pi r$. At $r = 10$: $V'' = -60\pi < 0$ ✓ (maximum).
4. Worked Example 3: Profit/Revenue#
A company’s profit (in rands) for selling $x$ items is given by $P(x) = -2x^2 + 400x - 5000$. Find the number of items that maximizes profit, and the maximum profit.
$P'(x) = -4x + 400$
Set $P'(x) = 0$: $-4x + 400 = 0 \Rightarrow x = 100$ items
$P(100) = -2(10000) + 400(100) - 5000 = -20000 + 40000 - 5000 = \text{R}15\,000$
Verify: $P''(x) = -4 < 0$ (always concave down = maximum) ✓
5. Rates of Change#
Some questions ask “At what rate is …” rather than “Find the maximum/minimum.” These are not optimization — they just want the derivative evaluated at a specific point.
Example#
A spherical balloon is inflated so that its radius increases at a rate of 2 cm/s. At what rate is the volume changing when the radius is 5 cm?
$V = \frac{4}{3}\pi r^3$
$\frac{dV}{dr} = 4\pi r^2$
At $r = 5$: $\frac{dV}{dr} = 4\pi(25) = 100\pi$
Since $\frac{dr}{dt} = 2$:
$\frac{dV}{dt} = \frac{dV}{dr} \times \frac{dr}{dt} = 100\pi \times 2 = 200\pi \approx 628.3 \text{ cm}^3\text{/s}$
6. Confirming Max vs Min#
After finding $f'(x) = 0$, always verify whether it’s a maximum or minimum:
| Method | Maximum | Minimum |
|---|---|---|
| Second derivative test | $f''(x) < 0$ | $f''(x) > 0$ |
| Context | Volume should be positive | Cost should be positive |
| Common sense | Check that the answer is physically reasonable |
🚨 Common Mistakes#
- Stopping at $x$: The question often asks for the maximum volume or minimum cost, not just the $x$-value. Substitute back into the original formula!
- Two-variable formulas: If your formula still has two unknowns when you try to differentiate, you missed the substitution step. Go back and use the constraint equation.
- Domain restrictions: Physical problems have restrictions ($x > 0$, $r < 6$, etc.). Reject solutions that violate these constraints.
- Units: Always include units in your final answer. If $V$ is in cm³, say so.
- Expansion errors: The substitution step often creates messy algebra. Take your time and double-check expansions before differentiating.
💡 Pro Tip: The “Constraint → Substitute → Differentiate” Mantra#
Every optimization problem with two variables follows this exact pattern. The constraint is usually a given total (surface area, perimeter, budget). Express one variable in terms of the other using the constraint, substitute into the formula you’re optimizing, then differentiate.
⏮️ Cubic Functions | 🏠 Back to Calculus