Worked examples
The question
Question: Two positive numbers multiply to 64.
Find the smallest possible sum.
1 Draw and name
Call the first number x, and the second y.
x = the first number
y = the second number
2 Write the sum
The sum is x plus y.
S = x + y
3 Use only one letter
They multiply to 64: x times y is 64.
xy = 64
So y = 64 over x.
y = 64x
Put it in place of y.
S = x + 64x
4 Which x make sense
The numbers are positive: x is more than 0.
x > 0 =
5 Slope = 0
The smallest value is where the slope is 0.
x gives 1.
64/x is 64x to the power −1, which gives −64/x².
S′(x) = 1 − 64x2
64x = 64x−1 → −64x−2
Set it to 0, and move the fraction across.
64x2 = 1
Times both sides by x², so they stay equal.
64 = x2
x squared is 64, so x is 8.
Minus 8 squared is also 64, but x must be positive.
x = 8 (not −8)
Careful!
The numbers must be positive.
Only x = 8 is allowed.
Careful: not x = −8 ✗
6 Make sure it is the smallest
Before 8, the slope is negative: going down.
After it, positive: going up.
Down, then up: the bottom.
A minimum.
S′(7) = −0.31 < 0 ↘, S′(9) = 0.21 > 0 ↗
7 Answer the question
The other number: 64/8 is 8.
The same number!
y = 648 = 8
So the smallest sum is 16.
S(8) = 8 + 8 = 16
8 Check
Let's check with x a bit smaller: 7 gives 16.14.
x = 7: S = 16.14
And x = 9 gives 16.11.
Both are bigger than 16.
Right!
x = 9: S = 16.11
Let's look back.
First: draw and name.
Then: write the sum.
Then: use only one letter.
Then: which x make sense.
Then: slope = 0.
Then: make sure it is the smallest.
Then: answer the question.
Last: check.
▶ Watch it solved
The question
Question: A rectangular field of 800 m² is fenced on three sides.
A river is the fourth side.
Find the least fencing needed.
1 Draw and name
The river is one side: no fence there.
The two sides that touch the river are x.
The side across is y.
x = each side touching the river
y = the side across
2 Write the fencing
The fence is two x sides and one y side.
L = 2x + y
3 Use only one letter
Two letters.
The area gives the link: x times y is 800.
xy = 800
So y = 800 over x.
y = 800x
Put it in place of y.
L = 2x + 800x
4 Which x make sense
A length is positive: x is more than 0.
x > 0 =
5 Slope = 0
The smallest value is where the slope is 0.
2x gives 2.
800/x is 800x to the power −1, which gives −800/x².
L′(x) = 2 − 800x2
800x = 800x−1 → −800x−2
Set it to 0, and move the fraction across.
800x2 = 2
Times both sides by x², so they stay equal.
800 = 2x2
Divide by 2.
x2 = 400
x squared is 400, so x is 20.
Minus 20 squared is also 400, but x must be positive.
x = 20 (not −20)
Careful!
A length cannot be negative.
Only x = 20 is allowed.
Careful: not x = −20 ✗
6 Make sure it is the smallest
Before 20, the slope is negative: going down.
After it, positive: going up.
Down, then up: the bottom.
A minimum.
L′(19) = −0.22 < 0 ↘, L′(21) = 0.19 > 0 ↗
7 Answer the question
The other side: 800/20 is 40.
y = 80020 = 40
So the least fencing is 80 m.
L(20) = 2 × 20 + 40 = 80 m
8 Check
Let's check with x a bit smaller: 19 gives 80.11.
x = 19: L = 80.11
And x = 21 gives 80.1.
Both are bigger than 80.
Right!
x = 21: L = 80.1
Let's look back.
First: draw and name.
Then: write the fencing.
Then: use only one letter.
Then: which x make sense.
Then: slope = 0.
Then: make sure it is the smallest.
Then: answer the question.
Last: check.
Answer L(20) = 2 × 20 + 40 = 80 m
▶ Watch it solved
The question
Question: A square sheet of card is 12 cm by 12 cm.
A square is cut from each corner and the sides are folded up to make an open box.
Find the largest volume.
1 Draw and name
Cut an x by x square from each corner.
The base that is left is 12 − 2x on each side.
x = the side of each cut square
Each side loses x at both ends: 12 minus x minus x.
12 − x − x = 12 − 2x
2 Write the volume
Volume is length times width times height.
V = length × width × height
The base is 12 − 2x by 12 − 2x, and the height is the fold, x.
V = (12 − 2x)(12 − 2x) × x
= x(12 − 2x)2
Multiply out the square first: 144, minus 48x, plus 4x².
(12 − 2x)2 = 144 − 48x + 4x²
Then times x.
Now we have V = 4x³ − 48x² + 144x.
= 4x3 − 48x2 + 144x
3 Which x make sense
x must be more than 0.
And 12 − 2x must be more than 0, so x is less than 6.
0 < x < 6 =
4 Slope = 0
The biggest value is where the slope is 0.
Use the power rule.
V′(x) = 4 × 3x^(3 − 1) − 48 × 2x^(2 − 1) + 144 × 1
Now we have V′(x) = 12x² − 96x + 144.
= 12x2 − 96x + 144
At a turning point the graph is flat for a moment: the slope is 0.
So we solve V′(x) = 0.
V′(x) = 0
12x2 − 96x + 144 = 0
Take out the 12.
12(x2 − 8x + 12) = 0
We need two numbers that multiply to 12 and add to −8: −2 and −6.
12(x − 2)(x − 6) = 0
−2 × (−6) = 12, −2 + (−6) = −8
A product is 0 only when one of its parts is 0.
a × b = 0 → a = 0 or b = 0
So x = 2 or x = 6.
x = 2 or x = 6
x = 6 is not allowed: the base would be 12 − 12 = 0.
No box at all.
x = 6: 12 − 2 × 6 = 0 ✗
Careful!
Check every x against the allowed ones.
6 cuts the whole sheet away.
Careful: not x = 6 ✗
So x = 2.
x = 2 ✓
5 Make sure it is the biggest
Before 2, the slope is positive: going up.
After it, negative: going down.
Up, then down: the top.
A maximum.
V′(1) = 60 > 0 ↗, V′(3) = −36 < 0 ↘
6 Answer the question
The base is 12 − 4 = 8.
So the largest volume is 128 cm³.
V(2) = 2 × 8 × 8 = 128 cm³
7 Check
Let's check with x a bit smaller: 1 gives 100.
x = 1: V = 100
And x = 3 gives 108.
Both are smaller than 128.
Right!
x = 3: V = 108
Let's look back.
First: draw and name.
Then: write the volume.
Then: which x make sense.
Then: slope = 0.
Then: make sure it is the biggest.
Then: answer the question.
Last: check.
Answer V(2) = 2 × 8 × 8 = 128 cm³
▶ Watch it solved
