Higher: optimise simple practical polynomial models
Build a practical model, use derivatives to locate candidates and compare them with the actual positive, finite or integer constraints.
Section 3.4 · Calculus (Higher only)
Exam scope and assessment notes
Higher only: Foundation has no section3.4 Calculus requirement. Higher also includes all shared Foundation content. These six original lessons teach variable/local versus interval rates, termwise integer-power differentiation, gradients/rates, stationary coordinates and graph-shape classification, one-dimensional kinematics and other simple practical models. H:3.4C requires both the gradient/rate and stationary-point lessons; H:3.4E requires both the motion and other-practical lessons with worked examples and all Guided, Independent and Transfer stages. The six-lesson split is StudyForA’s teaching design, not a syllabus-prescribed count. Small negative-integer powers with x≠0 conservatively retain integer-power breadth; fractional/trigonometric differentiation, integration and product/quotient/chain rules are not inferred requirements. H:3.4D classifies maxima and minima by general original-graph shape only; a second derivative is used for acceleration in motion, not as a substitute classification test. A derivative zero supplies an input, not the original stationary ordinate or an automatic turning point. Keep every original domain, physical restriction, admissible integer count and endpoint. Continuous relaxation supports differentiation for discrete counts; it does not make noninteger production feasible. Accepted boundary maxima need not be stationary. The separate H3.3D drawn-tangent construction is not replaced or newly credited by calculus. Fixed interactive examples support exact algebraic self-checks and approximate graph exploration; no general expression/root solver or written-work marker is supplied. Written work is not automatically marked. Attempt acknowledgements and revealed references do not award marks, prove a construction or establish mastery. All five Calculus source rows have structural teaching and staged-practice links; exhaustive subskill approval, specialist approval and cumulative formal assessment remain unestablished. The wider Maths A course and other Pearson qualifications remain incomplete. Publisher modular-map tags and uncorrected proof navigation are separate evidence: Book2 Graphs9 map pages386–415 differs from proof start385 and its later calculus subheadings. Full final-book and Teaching Hub body/pagination alignment remains unverified; chapter links confer no full-book or modular teaching credit.
Before you begin
- Form simple perimeter, area and volume expressions and expand products.
- Differentiate integer-power polynomials and solve simple equations.
- Use original graph shape and compare feasible model values with units.
Work on paper. Try each question before revealing help, and explain why your method works.
Quick readiness check
A rectangle has perimeter 18 m and sides x and y. Which area model and positive-side domain are correct?
What you will learn
- Derive area or volume as a single-variable polynomial from a practical constraint.
- Find stationary candidates and use original model shape to justify maxima.
- Apply positive-dimension, integer-count and boundary restrictions before choosing a practical answer.
f(x)=−x² + 14x. Original domain: x in (0, 14). Original values use cm²; first rates use cm²/cm.
The panels share the same numbered horizontal scale and have independent vertical scales. Original height, derivative value and derivative zero have different meanings. No input, derivative solution or reference point is preselected.
No input is preselected. Choose a point on your own drawing or a rate input you want to check.
Blue: original; purple: first derivative. Vertical scales differ. Filled boundary markers are included; open markers show excluded finite limits or holes. Traces split at original exclusions and are clipped to each panel; offscreen points are omitted rather than moved to a border. A wider view does not change a domain.
Derive the model and its admissible inputs
Start with the practical quantities and a constraint, rather than differentiating an unexplained expression. Express the required area, volume or profit using one variable, eliminate another variable if needed, then state the allowed inputs. Positive side lengths exclude degenerate rectangles or boxes even when their algebraic formula can still be evaluated.
Expand simple products into integer-power sums before differentiating. The derivative measures change in the output per change in the input: for area in cm² against a length in cm it has units cm²/cm, or cm. A stationary candidate comes from derivative zero, but that input must still satisfy the original practical conditions.
Compare a candidate with the original shape and boundaries
Return to the original model to calculate the requested maximum quantity and the corresponding dimensions or count. Use the original curve’s general shape to see whether the candidate is a hill rather than a valley. For a bounded domain, consider permitted endpoints as well; the greatest allowed value can occur at a boundary where the derivative is not zero.
Excluded degenerate endpoints are not valid designs. Their limiting output can still help describe the model’s shape, but do not report an excluded width or height as a usable solution. If a stationary candidate lies beyond a site restriction, optimise over the allowed interval instead of silently ignoring the extra constraint.
Treat discrete counts and model claims carefully
A production count may be restricted to integers even when a polynomial is written for it. Use the continuous polynomial as a candidate-finding aid, then compare the admissible integer counts and permitted boundaries. If the continuous candidate is not an integer, compare nearby allowed integers rather than assuming a rounded value must be best.
An invented profit, area or box model is only as broad as its stated assumptions. Show the original feasible model curve or its discrete points, retain units and label rejected candidates explicitly. A derivative calculation does not authorise negative dimensions, fractional items or an out-of-domain design, and it does not prove a real-world model beyond the simplified conditions supplied.
Sources for this lesson
Pause and explain
A profit model is allowed only at integer counts 0≤q≤10. Its continuous stationary candidate is q=4.4. What is the appropriate next step?
Worked example
A rectangle has perimeter 28 cm. Let one side length be x cm. Derive a one-variable area model and its positive-side domain, then use differentiation to find the maximum area and its dimensions. Draw the original area graph over the admissible interval and justify the maximum from its shape, respecting the excluded degenerate endpoints.
Show the worked solution
- If the other side is y cm, 2x+2y=28 gives y=14−x. The area is A=x(14−x)=14x−x² cm². Positive sides require 0<x<14; the inputs 0 and 14 would give degenerate rectangles and are excluded.
- Differentiate with x measured in cm: A′=14−2x, in cm² per cm. Set A′=0 to obtain x=7, which is admissible. The other side is 14−7=7 cm and the original area is A(7)=98−49=49 cm².
- The original model is the downward parabola A=49−(x−7)² on 0<x<14. It rises to the candidate then falls, with area tending to zero at either excluded end. Thus the square 7 cm by 7 cm gives the global maximum within the feasible domain; the derivative zero is the candidate, while the original shape justifies the maximum.
Answer A(x)=14x−x² cm² for 0<x<14. Maximum area 49 cm² at dimensions 7 cm by 7 cm. A′=14−2x, with derivative units cm²/cm.
Try a topic test
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Test this topic →From guided practice to a new situation
Write your answer and reasoning first. Use a hint only if you are stuck. The model solution is for self-checking; your written work is not automatically marked.
An invented profit model is P(q)=−2q²+48q−70 currency units for an integer production count 0≤q≤20. Use the continuous polynomial to find a candidate, then check admissibility, both neighbouring integer values and the permitted endpoints. Show the original model shape and state the maximum profit with its count.
Give me a hint
Differentiate the continuous extension as a candidate-finding aid. The actual decisions remain integer counts; substitute into P to compare profits.
Compare my reasoning
- For the continuous polynomial, P′(q)=48−4q, in currency units per item. Its zero gives q=12, a permitted integer. This derivative step does not change the original count restriction.
- Original profit values are P(11)=216, P(12)=218 and P(13)=216 currency units. The endpoints give P(0)=−70 and P(20)=90. Mark the admissible integer points on the original model rather than reporting a derivative as profit.
- The original downward parabola is P=218−2(q−12)², so all permitted counts have profit at most 218. The candidate and the comparisons agree: 12 items maximises the model profit. This is a conclusion about the invented polynomial on its stated count range.
Maximum profit 218 currency units at q=12 items. Neighbours: 216 at 11 and 13; endpoints: −70 at 0 and 90 at 20.
Look for these in your work
- Differentiate the continuous model but keep the integer decision domain.
- Compare original values at the candidate, neighbours and endpoints.
- Use the original downward shape to justify the maximum with correct units.
Cut equal squares of side x cm from each corner of an 18 cm square sheet and fold an open box. Derive V(x), state the admissible interval, and find the maximum volume using differentiation. Draw the original feasible volume curve, recover the box dimensions and reject any stationary value that gives a degenerate box.
Give me a hint
The height is x and each base side is 18−2x. Require every dimension to be positive before differentiating the expanded volume.
Compare my reasoning
- The volume is V=x(18−2x)²=4x³−72x²+324x cm³. Positive height and base sides require 0<x<9. The endpoints do not describe usable boxes.
- Differentiate: V′=12x²−144x+324=12(x−3)(x−9), in cm³ per cm. The algebraic candidates are x=3 and x=9. Reject x=9 because it is an excluded degenerate endpoint; x=3 is admissible.
- At x=3, the dimensions are height 3 cm and base 12 cm by 12 cm, giving V=3·12²=432 cm³. The original restricted volume curve rises to this hill then falls towards zero as x approaches 9; it also tends to zero as x approaches 0. This shape on the feasible interval supports the global maximum, without accepting the excluded stationary endpoint.
V=x(18−2x)² cm³ on 0<x<9. Maximum volume 432 cm³ at x=3 cm, with dimensions 3 cm by 12 cm by 12 cm. Candidate x=9 is excluded and degenerate.
Look for these in your work
- Derive the volume and strict positive-dimension interval.
- Find both algebraic stationary candidates and reject the inadmissible one.
- Recover the dimensions and original maximum volume and justify them from the feasible graph shape.
A rectangle against a straight wall uses 30 m of fence for the other three sides. Let the two equal perpendicular sides be x m and the far side y m. The site also requires 0<x≤6 m. Derive the area model and find the actual maximum area and dimensions. Explain why the unconstrained stationary point is not the answer and why the accepted maximum need not have derivative zero.
Give me a hint
Use 2x+y=30 and keep the site limit as well as positive dimensions. Inspect the original model only on the allowed interval, including its permitted boundary.
Compare my reasoning
- The fence constraint gives y=30−2x, so A=x(30−2x)=30x−2x² m². Combining positive dimensions with the site limit gives the actual domain 0<x≤6; throughout it the far side remains positive.
- The derivative is A′=30−4x, in m² per m. Its zero gives x=15/2=7.5 m, outside the site domain, so reject it as a practical design. The original downward parabola is still rising throughout the permitted interval, which ends before its unconstrained peak.
- The greatest allowed area is therefore at the included boundary x=6 m. Then y=30−12=18 m and A=6·18=108 m². At that boundary A′(6)=6 m² per m, not zero. A constrained boundary maximum does not have to be stationary; no larger allowed x is available.
Actual maximum 108 m² at x=6 m, y=18 m, on 0<x≤6. The unconstrained candidate x=7.5 m is rejected; the boundary maximum has A′(6)=6 m²/m.
Look for these in your work
- Derive the one-variable model and retain the site restriction.
- Reject the out-of-domain stationary candidate and inspect the feasible original shape.
- Give both dimensions and the boundary maximum, explaining its nonzero derivative.
Common mistakes
- Differentiating before deriving the constraint or omitting the admissible positive-dimension domain.
- Accepting an excluded stationary candidate or forgetting a permitted boundary that can supply the maximum.
- Treating an integer count as an unrestricted real decision, or reporting a derivative value instead of the original maximum quantity.
What can you explain now?
A continuous model’s stationary candidate is outside the permitted interval. Must the practical maximum occur where the derivative is zero?
Compare with the explanation
No. Compare the allowed model and its permitted boundaries.
A constraint can cut off the unconstrained peak, leaving the greatest admissible value at an endpoint where the derivative is nonzero.
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Make it stick
Explain why a derivative-zero candidate is followed by domain, boundary and original-output checks. Compare a positive-dimension design, an integer decision and a boundary-constrained optimum.
If you needed the solution, close it and solve the problem again from a blank page. A correct answer is stronger when you can explain the reason for each step.