If you can differentiate any expression but go blank when a question describes a box, a fence or a cost, the gap is usually modelling, not calculus. You have not yet turned the words into one formula in one variable. Once that formula exists, you already know what to do.
This page gives a repeatable way to build the model and a full worked example. It supports stationary points and optimisation.
What is actually going wrong?
There are three places where students commonly stall:
- No quantity named. The question asks for “the largest volume” but nothing is written as V = ….
- Two variables left. The formula has x and y, and there is nowhere to go.
- No check at the end. The stationary point is found, but nobody asks whether it is allowed, or whether it is a maximum or a minimum.
Each has a fix, and the routine below covers all three.
A routine that builds the model
- Draw and label. Give each length a letter. Mark what is fixed.
- Name the target. Write “V = …” or “A = …” for the quantity to maximise or minimise.
- Write the constraint. This is the fact that ties the variables together, such as a fixed total length.
- Eliminate to one variable. Use the constraint to write the target in terms of one letter.
- State the allowed range. Lengths are positive, and cutting cannot use more than the material you have.
- Differentiate, set to zero, solve.
- Justify and answer. Test maximum or minimum, reject values outside the range, and give the quantity asked for, with units.
Worked example
A square sheet of card has sides 12 cm. A square of side x cm is cut from each corner, and the sides are folded up to make an open box. Find the value of x that gives the greatest volume, and that volume.
Steps 1 and 2, draw and name the target. The base is a square of side (12 − 2x), because a length x is removed from each end. The height is x. So V = x(12 − 2x)².
Step 3, constraint. It is already built in: one variable, x, describes the whole box.
Step 5, allowed range. The base side must be positive, so 12 − 2x > 0, giving 0 < x < 6.
Step 6, differentiate. Using the product and chain rules:
dV/dx = (12 − 2x)² + x · 2(12 − 2x)(−2) = (12 − 2x)[(12 − 2x) − 4x] = (12 − 2x)(12 − 6x).
Set dV/dx = 0: 12 − 2x = 0 gives x = 6, and 12 − 6x = 0 gives x = 2.
Step 7, justify. x = 6 is outside the allowed range, because 0 < x < 6 excludes it, and it gives a box with no base, so V = 0. Keep x = 2.
Test the gradient either side: at x = 1, dV/dx = (10)(6) = 60 > 0. At x = 3, dV/dx = (6)(−6) = −36 < 0. The gradient goes from positive to negative, so x = 2 is a maximum.
Volume: V = 2 × (12 − 4)² = 2 × 64 = 128 cm³.
Independent check. At x = 1.9, V = 1.9 × 8.2² = 127.756. At x = 2.1, V = 2.1 × 7.8² = 127.764. Both are below 128, which agrees.
The mistake to watch for
A typical slip is to report both stationary values without checking the range.
Mistaken answer: “x = 2 or x = 6, so the maximum volume is at x = 6.”
The student stopped at the algebra. At x = 6 the base has side 0, so the “box” is flat and the volume is zero.
The correction is step 5: write the range before differentiating, and use it after. A stationary point is a candidate, not an answer. Maximum and minimum questions often have one extra root that the context rules out.
Check yourself
1. A farmer has 80 m of fencing for a rectangular pen. One side is an existing wall, so the fence covers the other three sides. Two sides have length x and the side parallel to the wall is y. Find x for the greatest area, and that area.
Show answer
Constraint: 2x + y = 80, so y = 80 − 2x. Area A = xy = 80x − 2x², with 0 < x < 40.
dA/dx = 80 − 4x = 0, so x = 20. The second derivative is −4, which is negative, so this is a maximum.
Then y = 40 and A = 800 m². Check: x = 19 gives 19 × 42 = 798, and x = 21 gives 21 × 38 = 798.
2. Two numbers add to 10. Find the two numbers that make the sum of their squares smallest, and that smallest sum.
Show answer
Let the numbers be x and 10 − x. S = x² + (10 − x)² = 2x² − 20x + 100.
dS/dx = 4x − 20 = 0, so x = 5. The second derivative is 4 > 0, a minimum. Both numbers are 5 and the smallest sum is 50. Check x = 4: 16 + 36 = 52.
3. An open box with a square base of side x cm and height h cm must hold 32 cm³. Find x that gives the smallest outer surface area, and that area.
Show answer
Volume constraint: x²h = 32, so h = 32/x². Surface area (base plus four sides) A = x² + 4xh = x² + 128/x.
dA/dx = 2x − 128/x² = 0, so x³ = 64 and x = 4. The second derivative 2 + 256/x³ = 6 > 0, a minimum. Then h = 2 and A = 16 + 32 = 48 cm². Check x = 3: 9 + 42.67 = 51.67. Check x = 5: 25 + 25.6 = 50.6.
What can you do next?
Try the calculus shape and rate explorer to see how the curve and tangent change as x moves, and the non-calculator working trainer to keep the arithmetic exact. Then practise more of the skill in the original practice hub.
When a story still gives no first line, it helps to have someone ask “what are you trying to make biggest?” at the right moment. That is what online one-to-one Additional Mathematics tuition can provide, and a paid trial is a practical way to find out whether the pace suits you.