4.8 - Maximum and minimum problems

Use calculus to choose the largest or smallest value that a problem actually allows. You will turn a geometric constraint into one expression, minimise positive reciprocal expressions and check domain boundaries as well as stationary points.

Maximising an area with a constraint

You have $48\,\mathrm{m}$ of fencing and a straight wall that provides one side of a rectangular enclosure. Would a square give the greatest area? The wall changes the problem: you pay for only three sides, so the best shape may surprise you.

You already know

An interior stationary point has zero first derivative. A negative second derivative there identifies a local maximum. Differentiate the quantity you want to optimise, then substitute into its original expression for the final value.

Let the two sides perpendicular to the wall each be $x$ metres, and the third fenced side be $l$ metres. The fencing constraint is $2x+l=48$ , so $l=48-2x$ . The area is $A=xl=x(48-2x)$ . Both dimensions must be positive, giving $0<x<24$ .

[DIAGRAM: asset_name: A rectangle against a wall; asset_slug: l2fm-fencing-constraint; caption: Only three sides use fencing, so the total fenced length is two short sides plus one long side; description: A rectangle lies against a horizontal fixed wall. The two perpendicular fenced sides are each labelled x metres; the opposite fenced side is labelled l metres. A note gives 2x plus l equals 48 and area A equals xl. The wall is clearly distinguished from the three sage fenced sides]
Only three sides use fencing, so the total fenced length is two short sides plus one long side

The constraint removes one variable. Once $l$ is written in terms of $x$ , you can use the differentiation you already know.

An interior optimum

For a differentiable quantity $Q(x)$ , an interior maximum or minimum must satisfy

\frac{\mathrm{d}Q}{\mathrm{d}x}=0

Keep only solutions in the problem's domain, justify the nature and substitute into $Q$ for the requested value. Compare included boundary values when the domain has endpoints.
Not on the formulae sheet: learn it.

In the enclosure problem the area is a downward-opening quadratic, so its one stationary point will be the maximum on the allowed interval.

Worked example 1: three fenced sides

Question. Use calculus to work out the dimensions and maximum area of the rectangular enclosure made with $48\,\mathrm{m}$ of fencing against a wall.

Plan. Differentiate the one-variable area, solve for $x$ and calculate the other dimension and area.

\begin{aligned} A&=48x-2x^2 &&\text{expand the area}\\ \frac{\mathrm{d}A}{\mathrm{d}x}&=48-4x &&\text{differentiate}\\ 48-4x&=0\quad\Longrightarrow\quad x=12 &&\text{stationary input}\\ \frac{\mathrm{d}^2A}{\mathrm{d}x^2}&=-4<0 &&\text{maximum}\\ l&=48-2(12)=24 &&\text{remaining fenced side}\\ A&=12\times24=288 &&\text{original area formula} \end{aligned}

Why this step? $x$ measures a side length. The question asks for area too, so a final answer of just $12$ would leave the optimisation unfinished.

Answer. The dimensions are $12\,\mathrm{m}$ by $24\,\mathrm{m}$ , giving maximum area $288\,\mathrm{m}^2$ .

Check. $2(12)+24=48$ , and $12$ lies in $0<x<24$ . The area tends to zero at either excluded end, so the interior maximum is also the greatest allowed area.

Worked example 2 changes only the amount of fencing to $64\,\mathrm{m}$ . Build the constraint again; do not change the optimum by guesswork.

Worked example 2: a longer fence

Question. Use calculus to work out the dimensions and maximum area when the same kind of enclosure uses $64\,\mathrm{m}$ of fencing.

Plan. Replace the total length in the constraint, then optimise the new area.

\begin{aligned} l&=64-2x,\qquad 0<x<32 &&\text{positive side lengths}\\ A&=x(64-2x)=64x-2x^2 &&\text{one-variable area}\\ \frac{\mathrm{d}A}{\mathrm{d}x}&=64-4x=0\quad\Longrightarrow\quad x=16 &&\text{stationarity}\\ \frac{\mathrm{d}^2A}{\mathrm{d}x^2}&=-4<0 &&\text{maximum}\\ l&=64-32=32 &&\text{remaining side}\\ A&=16\times32=512 &&\text{maximum area} \end{aligned}

Why this step? The two copies of $x$ both consume fencing, so the constraint contains $2x$ , not $x$ .

Answer. $16\,\mathrm{m}$ by $32\,\mathrm{m}$ , with maximum area $512\,\mathrm{m}^2$ .

Check. $16+16+32=64$ . Completing the square gives $A=512-2(x-16)^2$ , confirming that no allowed area exceeds $512$ .

The wall has a striking benefit. With the original $48\,\mathrm{m}$ , fencing all four sides gives $A=x(24-x)=144-(x-12)^2$ , at most $144\,\mathrm{m}^2$ . Letting the wall provide one side doubles the best possible area to $288\,\mathrm{m}^2$ , even though the amount of fencing is unchanged.

Watch out

Wrong: $x+l=48$ . There are two fenced sides of length $x$ , so the correct constraint is $2x+l=48$ . Count the actual sides before forming the expression.

The next enclosure changes the total again. Show the model before using calculus.

Minimising a positive reciprocal expression

In an expression containing both $x$ and $1/x$ , increasing $x$ makes one positive term larger and the other smaller, so the smallest total balances the two effects.

The domain is essential. In the examples below $x>0$ : a negative stationary solution is not an alternative answer, and $x=0$ is not defined at all.

Differentiating a reciprocal term

For constant $b$ and $x\ne0$ ,

\begin{aligned} \frac{b}{x}&=bx^{-1}\\ \frac{\mathrm{d}}{\mathrm{d}x}(bx^{-1})&=-bx^{-2}=-\frac{b}{x^2}\\ \frac{\mathrm{d}}{\mathrm{d}x}(-bx^{-2})&=2bx^{-3}=\frac{2b}{x^3} \end{aligned}

When $b>0$ and $x>0$ , the second derivative of this term is positive.
Not on the formulae sheet: learn it.

Write the negative power before differentiating, then keep the positive-domain restriction beside your solution.

Worked example 3: an exact fractional input

Question. For $x>0$ , $V=49x+\frac{81}{x}$ . Use calculus to show that $V$ has a minimum and work out its minimum value.

Plan. Solve the zero-gradient equation, retain its positive solution and test it before evaluating $V$ .

\begin{aligned} V&=49x+81x^{-1} &&\text{integer-power form}\\ \frac{\mathrm{d}V}{\mathrm{d}x}&=49-81x^{-2} &&\text{differentiate}\\ 49-\frac{81}{x^2}&=0\quad\Longrightarrow\quad49x^2=81 &&\text{multiply by }x^2\\ x&=\frac97 &&\text{retain the positive root}\\ \frac{\mathrm{d}^2V}{\mathrm{d}x^2}&=\frac{162}{x^3}>0 &&\text{throughout }x>0\\ V\left(\frac97\right)&=49\left(\frac97\right)+\frac{81}{9/7}=63+63=126 &&\text{evaluate the total} \end{aligned}

Why this step? The equation for $x^2$ gives two algebraic roots, but only $9/7$ is in the domain. A length-like positive input cannot be replaced by its negative root.

Answer. The minimum value is $126$ , attained at $x=\frac97$ .

Check. For $0<x<9/7$ , $49-81/x^2<0$ ; for $x>9/7$ , it is positive. Thus $V$ falls to this value and rises afterwards throughout the positive domain.

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4.8 - Maximum and minimum problems

Maximising an area with a constraint

Minimising a positive reciprocal expression

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