Optimization means finding a quantity’s largest or smallest value. For example, maximum area, minimum cost, or shortest distance are common examples. Calculus solves these critical points using optimization problems calculus.
Below you’ll find why the method works, a 6-step recipe, six solved problems from textbook classics to exam-level, the mistakes that cost the most marks, and practice problems with answers.
Why it works
Picture the graph of the quantity you want to maximize. At the top of a smooth hill, the curve is momentarily flat: the tangent line is horizontal, so the derivative is zero. The same is true at the bottom of a valley. That’s why the key step in solving optimization problems calculus is finding where \( f'(x) = 0 \).
The derivative being zero doesn’t guarantee a maximum, though. It only flags candidates. The Extreme Value Theorem fills in the rest for optimization problems calculus: a continuous function on a closed interval ( [a, b] ) always has an absolute maximum and minimum, and they can only occur at critical points or at the endpoints. So if you check every candidate, the largest value wins.
When the domain is open, like \( r > 0 \) for a can’s radius, there are no endpoints to check. Then you rely on the second derivative or on how the function behaves at the edges of the domain. If \( f'' > 0 \) everywhere (the graph is concave up, see concavity and inflection points), a single critical point must be the absolute minimum.
The 6-step method
- Read and draw. Sketch the situation and label variables.
- Write the objective: the formula for the quantity to maximize or minimize.
- Write the constraint linking the variables.
- Reduce to one variable by substituting the constraint into the objective.
- Differentiate, set equal to zero, solve.
- Check it’s really a max or min (second derivative test or endpoints) and answer the question asked, with units.
Example 1: largest fenced area
A farmer has 600 m of fence to enclose a rectangle along a river (no fence needed on the river side). What dimensions maximize the area?
- Objective: \( A = xy \), where \( x \) is the side perpendicular to the river.
- Constraint: \( 2x + y = 600 \Rightarrow y = 600 - 2x \).
- One variable: \( A(x) = x(600 - 2x) = 600x - 2x^2 \), with \( 0 < x < 300 \).
- \( A'(x) = 600 - 4x = 0 \Rightarrow x = 150 \).
- \( A''(x) = -4 < 0 \), so it’s a maximum.
Answer: 150 m by 300 m, area \( 45{,}000 \text{m}^2 \).
Example 2: the open box
First, cut equal squares of side \( x \) from the corners of a 12 × 12 cm sheet and fold up the sides. Which \( x \) gives the largest volume?
$$V(x) = x(12 - 2x)^2, \qquad 0 < x < 6$$
Expand first (or use the product rule with the chain rule):
$$V'(x) = 12x^2 - 96x + 144 = 12(x - 2)(x - 6)$$
The only critical point inside \( (0, 6) \) is \( x = 2 \). The volume is \( V(2) = 2\cdot 8^2 = 128 \text{cm}^3 \). At the endpoints \( V = 0 \), so this is the maximum.
Example 3: the cheapest can
A cylindrical can must hold 1000 cm³. What radius minimizes the surface area (the amount of metal)?
- Objective: \( ( S = 2\pi r^2 + 2\pi rh \) ).
- Constraint: \( \pi r^2h = 1000 \Rightarrow h = \frac{1000}{\pi r^2} \).
- One variable: \( S(r) = 2\pi r^2 + \frac{2000}{r} \).
- \( S'(r) = 4\pi r - \frac{2000}{r^2} = 0 \Rightarrow r^3 = \frac{500}{\pi} \Rightarrow r \approx 5.42 \text{cm} \).
Then \( h = \frac{1000}{\pi r^2} \approx 10.84 \) cm \( = 2r \). The best can is exactly as tall as it is wide.
Why is this a minimum? \(S''(r) = 4\pi + \frac{4000}{r^3} > 0\) for every r > 0. Thus, the graph is concave up on the domain. The single critical point is the absolute minimum. This is a common result for optimization problems in calculus.
Example 4: closest point on a curve
Find the point on \( y = \sqrt x \) closest to \( (3, 0) \).
Minimize the squared distance (same minimizer, simpler algebra):
$$D(x) = (x - 3)^2 + (\sqrt x)^2 = (x - 3)^2 + x$$
\( D'(x) = 2(x - 3) + 1 = 0 \Rightarrow x = \frac52 \). Therefore, the closest point is \( \left(\frac52, \sqrt{\frac52}\right) \approx (2.5, 1.58) \).
Example 5: open box with a fixed volume
A box with a square base and no top must hold \( 32{,}000 \text{cm}^3 \). Find the dimensions that use the least material.
- Objective: surface area \( S = x^2 + 4xh \) (base plus four sides), where \( x \) is the base side and \( h \) the height.
- Constraint: \( x^2h = 32{,}000 \Rightarrow h = \frac{32{,}000}{x^2} \).
- One variable: \( S(x) = x^2 + \frac{128{,}000}{x} \), with \( x > 0 \).
- \( S'(x) = 2x - \frac{128{,}000}{x^2} = 0 \Rightarrow x^3 = 64{,}000 \Rightarrow x = 40 \).
- \( S''(x) = 2 + \frac{256{,}000}{x^3} > 0 \), so it’s a minimum.
Answer: base 40 cm by 40 cm, height \( h = \frac{32{,}000}{1600} = 20 \) cm, using \( 4800 \text{cm}^2 \) of material. Also, the best open box is half as tall as it is wide.
Example 6: largest rectangle under a parabola
A rectangle has its base on the x-axis and its top corners on \( y = 12 - x^2 \). What’s the largest possible area?
By symmetry, put the corners at \( (\pm x, 12 - x^2) \). So the width is \( 2x \) and the height is \( 12 - x^2 \):
$$A(x) = 2x(12 - x^2) = 24x - 2x^3, \qquad 0 \le x \le \sqrt{12}$$
As a result, \( A'(x) = 24 - 6x^2 = 0 \Rightarrow x = 2 \) (the negative root is outside the domain). At the endpoints the area is 0, and \( A(2) = 4\cdot8 = 32 \).
Answer: the rectangle is 4 wide and 8 tall, with maximum area 32. This is the closed-interval method at work: compare the critical point with both endpoints.
Common mistakes
- Differentiating before reducing to one variable. \( A = xy \) has two unknowns. Use the constraint to eliminate one first.
- Ignoring the domain. Physical constraints (lengths > 0, a cut can’t exceed half the sheet) decide which critical points count, and endpoints can be the answer.
- Skipping the check. A critical point could be a max, a min or neither. Use the second derivative test or compare values.
- Answering the wrong question. If it asks for the maximum area, however, give the area, not just \( x \). Include units.
- Fighting square roots. For distance problems, minimize the distance squared. It has the same minimizer and much simpler algebra.
Where it’s used
Optimization is among calculus’s most practical parts. Businesses maximize profit when marginal revenue equals marginal cost. This means P'(x) = 0. Engineers minimize material, weight, or energy. Physics uses optimization problems in calculus. For example, light travels between two media. It takes the path of least time. Minimizing that time yields Snell’s law of refraction.
The setup skills overlap with related rates. Both translate words into an equation. However, optimization problems in calculus set a derivative to zero. Related rates solves for one. This is the key difference.
Solving optimization problems calculus with a calculator
Once you have the one-variable function, the gadget finds and classifies its critical points and checks the endpoints. For example, here’s Example 2 (the critical points calculator page explains the output in more detail):
Practice problems
Try these before checking the answers.
- Two positive numbers sum to 20; maximize their product
- Rectangle with perimeter 40 and maximum area
-
Minimize \( x + \frac{4}{x} \) for \( x > 0 \)
-
Open box from an 8 × 8 sheet by cutting squares of side \( x \) from the corners
- Point on \( y = x^2 \) closest to \( (0, 2) \)
Answers: (1) 10 and 10; (2) a 10 by 10 square; (3) \( x = 2 \), minimum value 4; (4) \( x = \frac43 \), volume \( \frac{1024}{27} \approx 37.9 \); (5) \( \left(\pm\sqrt{\frac32}, \frac32\right) \), at distance \( \frac{\sqrt7}{2} \approx 1.32 \).
FAQ
How do I know it’s a maximum and not a minimum?
Use the second derivative test, or compare values at the critical points and the endpoints of the domain. The second derivative calculator is handy for checking the sign of \( f'' \).
Why do we set the derivative equal to zero?
At a smooth peak or valley the tangent line is horizontal, so its slope, the derivative, is zero. Solving \( f'(x) = 0 \) finds every such point.
What’s the difference between a local and an absolute maximum?
A local maximum is the highest point in its neighborhood. The absolute maximum is the highest value on the whole domain. Optimization problems ask for the absolute one, so compare all candidates.
What if the derivative has no zeros?
Extreme values are at domain endpoints. They may not exist. The function may keep growing. Check points where the derivative is undefined. These also count as critical points for optimization problems in calculus.
Further reading
- Optimization (Paul’s Online Math Notes) — six worked problems and three ways to confirm you’ve found the optimum.
- More Optimization Problems (Paul’s Online Math Notes) — harder setups, like the longest pipe around a corner and the best trough angle.
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