P2 7.1 - Optimisation in context | Edexcel A-Level International Mathematics

Turn a practical constraint into a function of one variable, then use differentiation to find and justify an optimum. Keep the physical domain, allowed endpoints and requested quantity in view so that a stationary input becomes a useful answer.

Building a function with a domain

How should you shape an open box to hold a fixed volume using the least material? A wide base uses more material underneath but needs shorter walls. Calculus can balance those competing effects once the model has only one independent variable.

Start with a simpler trade-off: a rectangle whose perimeter is fixed. Write the quantity to optimise first, then use the constraint to remove the second variable.

Objective function and domain

An objective function is the quantity to maximise or minimise, expressed in terms of an input. Its domain is the set of allowed inputs. A physical length must be positive; any further restrictions in the question also belong in the domain.

Draw or list the actual sides or faces before forming a formula. A volume constraint and a surface-area objective measure different things: one cannot replace the other.

Worked example 1: a fixed perimeter

Question. A rectangle has perimeter $40\,\mathrm{m}$ and side lengths $x\,\mathrm{m}$ and $y\,\mathrm{m}$ . Express its area in terms of $x$ , with the domain for a nondegenerate rectangle.

Plan. Use the perimeter equation to eliminate $y$ from $A=xy$ .

\begin{aligned} 2x+2y&=40\implies y=20-x\\ A(x)&=x(20-x)=20x-x^2 \end{aligned}

Both sides must be positive: $x>0$ and $20-x>0$ .

Answer. $A=20x-x^2$ square metres, for $0<x<20$ .

Check. At $x=6$ , $y=14$ : the perimeter is $2(6+14)=40$ and the area is $84$ , also given by $20(6)-6^2$ .

Change only which sides need fencing: a straight river supplies one side, so the same $40\,\mathrm{m}$ of fencing now covers three sides.

Worked example 2: one side supplied by a river

Question. A rectangular enclosure beside a straight river uses $40\,\mathrm{m}$ of fencing for its two widths $x$ and its length $y$ . Find its area function and physical domain.

Plan. Count the fenced sides, then eliminate the length from the area formula.

\begin{aligned} 2x+y&=40\implies y=40-2x\\ A(x)&=x(40-2x)=40x-2x^2\\ x>0,\quad40-2x>0&\implies0<x<20 \end{aligned}

Answer. $A=40x-2x^2$ square metres, for $0<x<20$ .

Check. At $x=6$ , $y=28$ . The fenced sides total $6+6+28=40$ , and $6\times28=168=40(6)-2(6)^2$ .

Endpoints $x=0$ and $x=20$ would collapse an enclosure to zero area; they are excluded from these physical domains. A different problem may explicitly allow endpoints, for example an operating temperature in a closed range.

Watch out

Do not use $2x+2y=40$ for the river enclosure. The constraint must count the sides actually present in the question.

Build the same three-side model with a new fence length.

Finding candidates

Differentiation finds possible interior optima; the domain decides which of those candidates are allowed.

For a differentiable function, an interior maximum or minimum must have zero derivative. Thus solve $F'(x)=0$ , but retain only solutions in the domain. Allowed endpoints must be considered separately. All the objective functions here are differentiable inside their stated domains.

A denominator is often easiest to differentiate as a negative power. For example, $108/x=108x^{-1}$ , with derivative $-108x^{-2}$ .

Worked example 3: a reciprocal term

Question. A surface-area model is $S=2x^2+108/x$ , where $x>0$ . Find the stationary input.

Plan. Differentiate as a sum of powers, then clear the denominator in the zero-gradient equation.

\begin{aligned} \frac{\mathrm{d}S}{\mathrm{d}x}&=4x-108x^{-2}\\ 4x-\frac{108}{x^2}&=0\\ 4x^3-108&=0 &&\text{multiply by }x^2>0\\ x^3&=27\implies x=3 \end{aligned}

Answer. The stationary input is $x=3$ , which is in the domain.

Check. At $3$ , the derivative is $12-108/9=0$ . Nearby gradients at $2$ and $4$ are $-19$ and $9.25$ , consistent with a possible minimum.

Change only the allowed domain. An algebraically correct root can cease to be a candidate for the problem.

Worked example 4: a stationary input outside the domain

Question. The same model $S=2x^2+108/x$ is restricted to $4\le x\le8$ . Find its stationary candidates in this interval.

Plan. Solve the same derivative equation, then apply the interval restriction.

\frac{\mathrm{d}S}{\mathrm{d}x}=0 \implies x^3=27\implies x=3.

But $3\notin[4,8]$ .

Answer. There are no stationary candidates in the allowed interval. The endpoints $4$ and $8$ still need comparison when finding the optimum.

Check. For $x\ge4$ , $4x\ge16$ and $108/x^2\le6.75$ , so $S'(x)>0$ . There is no zero derivative in the interval.

Multiplying by $x^2$ was valid because $x=0$ was already excluded. Solving $x^3=27$ gives $3$ , not $\pm3$ : $(-3)^3=-27$ .

Watch out

A stationary value of the objective and the input producing it are different answers. Here $x=3$ is an input; $S(3)=54$ is the associated surface area.

Try a new reciprocal coefficient and check the permitted interval.

Checking endpoints and type

A local minimum answers a question about nearby inputs; a global minimum is the least value over the whole allowed domain.

For a continuous function on a closed interval, compare the objective at every stationary input in the interval and at both endpoints. The smallest value is the global minimum; the largest is the global maximum. A second derivative test can classify an interior point, but it does not replace the endpoint comparison.

Worked example 5: an interior minimum and an endpoint maximum

Question. Find the greatest and least values of $F(x)=x^2-6x+11$ for $0\le x\le5$ .

Plan. Find the stationary input, check its type and compare its value with the endpoint values.

F'(x)=2x-6=0\implies x=3,\qquad F''(x)=2>0.

Candidate $x$	$0$	$3$	$5$
$F(x)$	$11$	$2$	$6$

Answer. Minimum $2$ at $x=3$ ; maximum $11$ at $x=0$ .

Check. Completing the square gives $F(x)=(x-3)^2+2$ . The minimum is visibly $2$ ; of the allowed endpoints, zero is further from $3$ and gives the larger value.

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P2 7.1 - Optimisation in context

Building a function with a domain

Finding candidates

Checking endpoints and type

Continue this lesson.

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