Optimisation | CAIE 9709 Pure Mathematics 1 Notes

Pure Mathematics 1

Mathematics9709

Choose the quantity to make greatest or least

Optimisation means finding the greatest or least possible value under given restrictions. The target might be area, volume or cost. A constraint is the fixed relationship restricting your choices.

Only the two x sides and the L side use fencing: 2x + L = 50.

Worked example

Use 50 m of fencing for three sides of a rectangular enclosure against a wall. Maximise its area.

Let each side perpendicular to the wall be $x$ metres, and the remaining fenced side be $L$ metres. The wall needs no fence, so count only three sides:

\begin{gathered}2x+L=50\\L=50-2x\end{gathered}

The target is rectangular area, length times width: $A=xL$ . Substitute the constraint to express it using one variable. As $x$ changes, $L$ changes too; it is not a constant.

\begin{aligned}A&=x(50-2x)\\&=50x-2x^2\end{aligned}

Both lengths must be positive: $x>0$ and $50-2x>0$ . The physical domain is $0<x<25$ .

Solve, justify and answer the requested quantity

\begin{gathered}A'=50-4x=0\\x=12.5\end{gathered}

This value is allowed. Since $A''=-4<0$ , it gives a maximum. Also, $A'$ is positive before 12.5 and negative after it throughout the physical domain, so this is the greatest area, not just a local maximum.

The other length is $L=25$ m. The area is $12.5\times25=312.5$ m². The value of $x$ alone would not answer an area question.

Your turnPractice

The same arrangement now uses 60 m of fencing. Find the largest area and both dimensions. Justify that your answer is a maximum.

Show worked answer

$L=60-2x$ , $A=60x-2x^2$ , $0<x<30$ . Setting $A'=60-4x=0$ gives $x=15$ , $L=30$ . Area $450\text{ m}^2$ ; $A''=-4<0$ , with rise then fall over the domain.

Restrictions can put the optimum at an endpoint

On a closed interval, both endpoints are allowed. If the model has no breaks and a derivative throughout the interior, compare their target values with the valid stationary values. A boundary optimum need not have zero derivative.

Worked example

In the 50 m problem, the site additionally requires 1 ≤ x ≤ 10.

The stationary input 12.5 is outside the interval. The area is increasing throughout $1\le x\le10$ , so the greatest area occurs at $x=10$ : $A=300\text{ m}^2$ .

If an endpoint is excluded, its value cannot be attained. Check what the model approaches there; do not automatically report it as an actual maximum or minimum.

Your turnPractice

The 50 m enclosure instead requires $15\le x\le20$ . Find its greatest area. Is there an allowed stationary point?

Show worked answer

No: 12.5 is outside the interval. $A'<0$ throughout, so choose the left endpoint $x=15$ . Then $L=20$ and $A=300\text{ m}^2$ .

Choose the quantity to make greatest or least

Optimisation means finding the greatest or least possible value under given restrictions. The target might be area, volume or cost. A constraint is the fixed relationship restricting your choices.

Only the two x sides and the L side use fencing: 2x + L = 50.

Worked example

Use 50 m of fencing for three sides of a rectangular enclosure against a wall. Maximise its area.

Let each side perpendicular to the wall be $x$ metres, and the remaining fenced side be $L$ metres. The wall needs no fence, so count only three sides:

\begin{gathered}2x+L=50\\L=50-2x\end{gathered}

The target is rectangular area, length times width: $A=xL$ . Substitute the constraint to express it using one variable. As $x$ changes, $L$ changes too; it is not a constant.

\begin{aligned}A&=x(50-2x)\\&=50x-2x^2\end{aligned}

Both lengths must be positive: $x>0$ and $50-2x>0$ . The physical domain is $0<x<25$ .

Solve, justify and answer the requested quantity

\begin{gathered}A'=50-4x=0\\x=12.5\end{gathered}

The other length is $L=25$ m. The area is $12.5\times25=312.5$ m². The value of $x$ alone would not answer an area question.

Your turnPractice

The same arrangement now uses 60 m of fencing. Find the largest area and both dimensions. Justify that your answer is a maximum.

Show worked answer

$L=60-2x$ , $A=60x-2x^2$ , $0<x<30$ . Setting $A'=60-4x=0$ gives $x=15$ , $L=30$ . Area $450\text{ m}^2$ ; $A''=-4<0$ , with rise then fall over the domain.

Restrictions can put the optimum at an endpoint

Worked example

In the 50 m problem, the site additionally requires 1 ≤ x ≤ 10.

The stationary input 12.5 is outside the interval. The area is increasing throughout $1\le x\le10$ , so the greatest area occurs at $x=10$ : $A=300\text{ m}^2$ .

If an endpoint is excluded, its value cannot be attained. Check what the model approaches there; do not automatically report it as an actual maximum or minimum.

Your turnPractice

The 50 m enclosure instead requires $15\le x\le20$ . Find its greatest area. Is there an allowed stationary point?

Show worked answer

No: 12.5 is outside the interval. $A'<0$ throughout, so choose the left endpoint $x=15$ . Then $L=20$ and $A=300\text{ m}^2$ .