Solve a maximum or minimum problem by differentiation

A rectangle has side lengths x and y cm with a fixed sum of 20 cm. What is the first reduction for maximising its area?

Change the cut size and compare all five quantities. Find where increasing the cut ceases to increase volume

Why can a larger cut reduce the capacity?

Optimisation with a constraint

An optimisation problem asks for the largest or smallest value of an objective. Use the constraint to write that objective in one variable, then check the full practical domain

1. 1Name the objective Write the quantity to be made largest or smallest before using the constraint \(\displaystyle A=xy\)
2. 2Use the constraint Rearrange the constraint to express one variable in terms of the other, then substitute into the objective \(\displaystyle x+y=20\quad\Rightarrow\quad y=20-x\quad\Rightarrow\quad A(x)=x\left(20-x\right)\)
3. 3State the practical domain For a nondegenerate rectangle, both side lengths are positive: 0<x<20. Extend the area formula to [0,20] only to compare the zero-area boundary limits \(\displaystyle 0<x<20\)
4. 4Find stationary candidates Set the derivative equal to zero and solve for candidates in the practical domain \(\displaystyle A'(x)=20-2x=0\quad\Rightarrow\quad x=10\)
5. 5Evaluate the stationary area Substitute the stationary side length into the objective \(\displaystyle A(10)=10(20-10)=100\)
6. 6Check the lower boundary extension At x=0, the extended area formula gives the zero-area limit of a degenerate shape \(\displaystyle A(0)=0(20-0)=0\)
7. 7Check the upper boundary extension and compare The upper boundary extension also gives zero area. Comparing both limits with the stationary area shows that 100 cm² is the attained maximum \(\displaystyle A(20)=20(20-20)=0\)

Given

\(\displaystyle A'(x)=20-2x=0\)

Solve the stationary equation for the side length x.

Given

\(\displaystyle A(10)=10\left(20-10\right)\)

Evaluate the area at the stationary side length

With x+y=20 and both side lengths varying, which line prepares the one-variable area objective?

A prototype tray investigation

A workshop compares tray designs folded from a 48 cm by 30 cm rectangular board after equal squares of side x cm are removed from all four corners. The model ignores thickness, tabs and folding losses. We first maximise capacity with a continuous cut size, then require half-centimetre cuts and limit the removed area. These are stated modelling assumptions, not measurements of a manufactured tray

Form the model from dimensions

Let x be the cut size and tray height. Each base dimension is reduced by twice the cut size. Positive dimensions set the practical domain. Capacity is the product of height and base dimensions in cm³; removed area is measured in cm². Both height and base area change, so selecting the tallest tray alone cannot solve the question

Model

\(\displaystyle 30-2x>0\)

What excluded upper boundary does positive base width impose on x?

Identify the global continuous maximum

Differentiate the whole capacity rule and factor. Only the stationary input 6 lies inside (0,15); the other stationary input 20 violates the domain. The derivative is positive before 6 and negative after it within the domain. Volume tends to zero at both boundaries, so the interior maximum is global for this model

Model

\(\displaystyle V=x(L-2x)(W-2x),\quad x=6,\quad L-2x=36,\quad W-2x=18\)

Calculate the model capacity at the valid stationary input

Compare capacity with waste

A 3 cm cut produces 3024 cm³ and removes 36 cm², or 2.5% of the board. A 6 cm cut produces 3888 cm³ and removes 144 cm², or 10%. The larger capacity uses four times as much removed area. A capacity-only optimum does not settle a decision that also limits waste

Model

\(\displaystyle w=4(6)^2/(48\times30)\)

Express the removed fraction for x=6 as an exact fraction

Apply a practical restriction before rounding

Suppose the removed area may not exceed 8% of the board and cuts must be positive multiples of 0.5 cm. The continuous waste bound is shown below. Rounding the bound up to the next half-centimetre would violate it. Among permitted feasible cuts, volume is still increasing, so the largest feasible cut of 5 cm is optimal: its capacity is 3800 cm³, with the removed fraction shown below

Model

\(\displaystyle x=n/2,\quad n^2\le(8/100)(48)(30),\quad n\in\mathbb{N};\qquad V(x)=x(48-2x)(30-2x),\quad V'(x)>0\text{ for }0<x<6\)

What permitted cut maximises capacity under the 8% cap?

An exact finite search with a feasibility test

For integer board dimensions and half-centimetre cuts, use the cut size, cap and exact feasibility test shown below. This avoids accepting a cut because a rounded waste percentage looked small enough. Enumerate all geometric candidates, retain feasible ones, compare exact volumes and retain ties. Report no feasible design if none survives

1. 1Sequence and repeat Set bestVolume to no value and bestCuts to empty. For each integer n from 1 through min(L,W)−1, set x=n/2 and compute both base dimensions
2. 2Decide and compare If the feasibility inequality holds, compute capacity from the three dimensions. Replace the best record when capacity is larger; append the cut on a tie. After the final candidate, report the retained cuts, capacity and actual removed fraction \(\displaystyle n^2q\le pLW\quad\Rightarrow\quad V=x(L-2x)(W-2x)\)
3. 3Worked checks For 48×30 and p/q=2/25, n=10 passes because 2500≤2880, while n=11 fails because 3025>2880. The best capacity is 3800 cm³ at x=5.

Generalise before comparing boards

For positive L≥W, the same geometry gives the cubic below. Solving the derivative quadratic gives the smaller stationary root x*. It lies inside the physical interval and gives the continuous maximum. A waste cap may force a smaller cut, and permitted increments require a separate finite search. The general formula is a model result, not permission to ignore either restriction

Use different checks for the same conclusion

For the worked 48×30 model: direct dimensions at 6 give 3888; derivative signs establish the maximum; a separately enumerated half-centimetre table confirms 6 without the cap and 5 under the 8% cap; the original geometry excludes cuts at or above 15. A graph can locate the maximum but cannot alone certify the exact cut or feasibility. Units distinguish cm, cm² and cm³

A model report with contextual limits

Under the thin-board assumptions, the continuous capacity maximum for a 48 cm by 30 cm sheet is 3888 cm³ at a 6 cm cut. It satisfies the half-centimetre increment but removes 10% of the board. With an 8% cap, a 5 cm cut gives the largest feasible capacity, 3800 cm³, and removes about 6.944%. The cap therefore reduces model capacity by 88 cm³. I checked dimensions, derivative signs and an exact feasible-cut table. I recommend 5 cm only under these stated restrictions; thickness, tabs, strength and assembly losses would need a prototype check before manufacture

Your investigation: define the question and model

Choose one board from 36×24, 42×30 or 48×30 cm. Choose two removed-area caps from 6%, 10% and 14%, and explain why comparing them is a useful question. Require positive half-centimetre cuts. Before calculating, specify variables, units, assumptions, constraints, relevant mathematics and an objective; derive your own dimension, capacity and waste rules. Predict which restriction might be active. Sketch the proposed cut-and-fold geometry on paper

1. 1Formulation State a precise question, relevant mathematics, variables and units, chosen parameters, assumptions, constraints and the objective. Derive the rule from the geometry; explain how the domain follows from positive dimensions
2. 2Analysis and algorithm Separate the continuous model from permitted half-centimetre cuts. Show derivative signs and boundary reasoning, then give the ordered candidate loop, feasibility test, comparison and stopping condition. Retain actual technology settings/output if you use technology
3. 3Verification Use at least three distinct checks: original dimensions/substitution, derivative signs or completing a symbolic check, an independently enumerated feasible table/plot, and units/boundary checks. Reusing one rounded result in three places is not three independent checks
4. 4Inference and communication Compare at least two alternatives numerically, state the recommendation in ordinary language with units and limitations, and distinguish a mathematical maximum from a tested physical design. Explain what changes when the cap or assumptions change
5. 5Written and external work is ungraded Keep your formulation, drawings, algorithm record, calculations, checks and report together

Your investigation: analyse, adapt and verify

Find the unconstrained continuous maximum with calculus, including rejected stationary points and boundary reasoning. Adapt the supplied finite-search procedure to both chosen caps; retain your exact feasible table or actual technology record and settings. Compare the continuous maximum with each permitted optimum. Verify the conclusions by at least three distinct methods. Extend your question: for a target capacity you justify, find the smallest-waste permitted design, or explain why none is feasible

1. 1Formulation State a precise question, relevant mathematics, variables and units, chosen parameters, assumptions, constraints and the objective. Derive the rule from the geometry; explain how the domain follows from positive dimensions
2. 2Analysis and algorithm Separate the continuous model from permitted half-centimetre cuts. Show derivative signs and boundary reasoning, then give the ordered candidate loop, feasibility test, comparison and stopping condition. Retain actual technology settings/output if you use technology
3. 3Verification Use at least three distinct checks: original dimensions/substitution, derivative signs or completing a symbolic check, an independently enumerated feasible table/plot, and units/boundary checks. Reusing one rounded result in three places is not three independent checks
4. 4Inference and communication Compare at least two alternatives numerically, state the recommendation in ordinary language with units and limitations, and distinguish a mathematical maximum from a tested physical design. Explain what changes when the cap or assumptions change
5. 5Written and external work is ungraded Keep your formulation, drawings, algorithm record, calculations, checks and report together

Model

\(\displaystyle V=(11/2)(42-11)(30-11)\)

For the 42×30 comparison, independently calculate the capacity at a 5.5 cm cut

Your investigation: communicate a recommendation

Write a short report with a labelled model diagram, derived formulas, domain/constraint statement, algorithm or exact table, comparison of at least two feasible alternatives, varied checks and a justified recommendation. Explain in everyday language what is gained or lost by relaxing the cap. Discuss assumptions and distinguish calculated model output from any measured evidence. Include the extension question and its conclusion. Review the complete work against every criterion, then annotate any correction

1. 1Formulation State a precise question, relevant mathematics, variables and units, chosen parameters, assumptions, constraints and the objective. Derive the rule from the geometry; explain how the domain follows from positive dimensions
2. 2Analysis and algorithm Separate the continuous model from permitted half-centimetre cuts. Show derivative signs and boundary reasoning, then give the ordered candidate loop, feasibility test, comparison and stopping condition. Retain actual technology settings/output if you use technology
3. 3Verification Use at least three distinct checks: original dimensions/substitution, derivative signs or completing a symbolic check, an independently enumerated feasible table/plot, and units/boundary checks. Reusing one rounded result in three places is not three independent checks
4. 4Inference and communication Compare at least two alternatives numerically, state the recommendation in ordinary language with units and limitations, and distinguish a mathematical maximum from a tested physical design. Explain what changes when the cap or assumptions change
5. 5Written and external work is ungraded Keep your formulation, drawings, algorithm record, calculations, checks and report together

One complete comparison to check your reasoning

For a 42 cm by 30 cm board, the continuous maximum is about 3244.44 cm³ at the cut shown below. Exact half-centimetre search gives the cuts and capacities shown for the 6%, 10% and 14% caps, respectively. Their removed fractions are shown in the same order. Moving from the 10% cap to the 14% cap gains only 0.5 cm³ in this grid model. A recommendation can favour the stricter cap if that small gain is not worth the extra removed area; its justification must state the decision criterion

1. 1Extension example If the chosen target is 3200 cm³ on this 42×30 board, a 5 cm cut gives exactly 3200 cm³, while smaller cuts fall short. Larger cuts that meet the target remove more area. The minimum-waste permitted choice is therefore 5 cm. Under a 6% cap that target is infeasible. Verify both statements with your complete feasible table
2. 2Written and external work is ungraded Keep your formulation, drawings, algorithm record, calculations, checks and report together