NCEA Level 1Optimal solutionsAS 91947

Optimal solutions: break-even and maximum area

What this lesson covers

The best option depends on the quantity being compared, so the comparison has to be set up before the choice is made.
Where two cost rules cross is the point at which the better option changes.

Key idea

Optimal means best at something

Cheapest, largest and quickest are different questions.
So name the measure before you compare anything.
Then the optimal choice is the best by that measure.

Situation	Measure	Best is
Hiring a kayak	total cost	the lowest
Fencing a garden	area enclosed	the largest
Getting to school	time taken	the shortest
So first	name the measure	then compare

Key idea

Two hire plans, as equations

Each plan is a fixed fee plus a rate for every hour.
The rate multiplies h. The fixed fee stands alone.
P = 10 + 2h and Q = 4 + 3h.

Plan	In words	Equation
Plan P	$10, then $2 an hour	P = 10 + 2h
Plan Q	$4, then $3 an hour	Q = 4 + 3h
h	hours of hire	whole hours
Fixed fee	the number on its own	rate goes with h

Worked example

Try a few hires first

Short hires: Q wins, because its fee is small.
Long hires: P wins, because its rate is small.
So somewhere in between, the winner changes.

Hire	Plan P	Plan Q
2 hours	$14	$10, Q wins
4 hours	$18	$16, Q wins
8 hours	$26	$28, P wins
Somewhere between	the two	are equal

Worked example

The break-even point

Break-even means the two costs are equal.
Set P equal to Q and solve like any equation.
h = 6, and both plans cost $22. Checked.

Step	Working	Check
P = Q	10 + 2h = 4 + 3h
Take 2h off both	10 = 4 + h
Take 4 off both	h = 6
Check	P: 10 + 12 = 22	Q: 4 + 18 = 22

Worked example

Where the two plans cross

Q starts lower but climbs faster.
They meet at 6 hours and $22.
After the crossing, P is the lower one.

Worked example

When is Plan P cheaper?

Cheaper means less than, so write P < Q.
Solve it exactly as you solved the equation.
h > 6. No negative multiplier, so no reversal.

Step	Working	Result
P cheaper	10 + 2h < 4 + 3h
Take 2h off both	10 < 4 + h
Take 4 off both	6 < h	so h > 6
Check h = 7	P $24, Q $25	P cheaper

Worked example

Answer in hours, not in h

Hire is charged in whole hours.
So h > 6 means seven hours or more.
Give all three cases, including the tie.

Hire	Better plan	Because
1 to 5 hours	Q is cheaper	h < 6
Exactly 6 hours	same cost, $22	h = 6
7 hours or more	P is cheaper	h > 6
Whole hours	so h > 6 means	7, 8, 9, ...

Worked example

Justifying the choice

State the choice and the measure.
Give the numbers that prove it.
Say why, using the break-even point.

Part	For a 3-hour hire
The choice	Plan Q, for 3 hours
The measure	it is the cheaper plan
The numbers	Q costs $13, P costs $16
The reason	Q is cheaper for any hire under 6 hours

Key idea

A different measure: the most area

12 m of fence, with a wall along one side.
The width x is fenced twice, so it uses 2x.
Area is width times length: A = x(12 − 2x).

Part	Reasoning	Expression
Fence	12 m, wall on one side
Width x	used twice	2x
Length	what is left	12 − 2x
Area	width × length	A = x(12 − 2x)

Worked example

Try some widths

Too narrow is small. Too wide is small too.
The areas rise, then fall in the same steps.
That mirror pattern is the clue.

Width	Length	Area
1 m	10 m	10 m²
2 m	8 m	16 m²
3 m	6 m	18 m²
4 m	4 m	16 m²
5 m	2 m	10 m²

Worked example

Where the area is zero

The model is already factorised.
A product is zero when one factor is zero.
So the zeros are x = 0 and x = 6.

Case	Working	Meaning
A = 0 when	x = 0	no width
or when	12 − 2x = 0	x = 6
x = 6	length is 0	no garden
Zeros	0 and 6	the ends

Worked example

The maximum is halfway

A parabola is symmetric, and this one opens downward.
Halfway between 0 and 6 is x = 3.
A = 3 × 6 = 18 m². That is the maximum.

Worked example

Any length of fence

Fence L metres: x(L − 2x), which opens downward.
The zeros are 0 and L/2, so the middle is L/4.
Maximum area: L²/8. With L = 12 it is 18.

Step	Working	Result
Fence L	A = x(L − 2x)
Zeros	x = 0 and x = L/2
Midpoint	x = L/4	length L/2
Maximum	(L/4)(L/2)	L²/8
Check L = 12	144 ÷ 8	18, as before

Summary

What to take away

Seven lines, and every one starts by naming the measure.


Optimal	best by a measure you name first
Two plans	write an equation for each
Break-even	set them equal and solve
Which is cheaper	an inequation, read in context
Whole numbers	h > 6 means 7 hours or more
Maximum of x(L − 2x)	halfway between the zeros
Justify	choice, measure, numbers, reason

Your turn

Your turn: four questions

Pause here and write each one out in full.
Question 2 wants a whole number of guests.
Question 4 has four sides fenced, not three.

	Question
1	Hall P $120 + $5/guest, Q $40 + $9/guest. Equal cost when?
2	Use an inequation: for how many guests is Hall P cheaper?
3	A garden: A = x(16 − 2x) (x in metres). Find its maximum area.
4	20 m of fence goes round all 4 sides. Width for largest area?

Show answer

1. 20 guests. 120 + 5n = 40 + 9n gives 80 = 4n, so n = 20, and both halls cost $220. 2. 21 guests or more. 120 + 5n < 40 + 9n gives 80 < 4n, so n > 20, and guests come in whole numbers. Check 21: P is $225, Q is $229. 3. 32 m². Expanded, A = −2x² + 16x; the x² term is negative, so the graph opens downward and its turning point is a maximum. The zeros are x = 0 and x = 8, the midpoint is x = 4, the length is 16 − 8 = 8, and 4 × 8 = 32. 4. 5 m. Width x leaves length 10 − x, so A = x(10 − x) with zeros 0 and 10. Halfway is 5, a 5 m square of 25 m², and widths 4 and 6 give only 24 m².

Worked example

The answers

20 guests, where both halls cost $220.
21 guests or more.
Opens downward: 32 m², at a width of 4 m.
A width of 5 m: a square of 25 m².

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