Since the perimeter of the top surface is 22 cm,

L + B + L + B = 22
2 (L + B) = 22
L + B = 11

The phrase “144 one-cm cubes” just simply means that the values of L, B and H have to be whole numbers and not decimals. Also, each of L, B and H must be greater than 2.

What are the possible combos?

L + B = 11

L and B cannot be 1 or 2 since the sides exceed 2 cm.

Can L and B be 8 and 3? Possible, because 144 / (8 x 3) = 6 which is a whole number.

Can L and B be 7 and 4? No, because 144 / (7 x 4) is not a whole number.

Can L and B be 6 and 5? No, because 144 / (6 x 5) is not a whole number.

This means that L and B must be 8 or 3, so H has to be 6.

Volume of cuboid, V = 144cm³

Perimeter of top = Length(L) + Breadth(B)

2L + 2B = 22cm

L + B = 11cm


Since V = 144cm², perform prime factorisation (sec 1 basics) to see what combinations of terms you can get.

144 = 2⁴ × 3² (or 2 × 2 × 2 × 2 × 3 × 3)


Since L,B,H > 2,

Then we need to combine some of the 2's together


144
= 2 × 2 × 2 × 2 × 3 × 3
= 2³ × 3 × (2 × 3)
= 8 × 3 × 6

8 + 3 = 11 so L = 8cm, B = 3cm, H = 6cm


Using prime factorisation, you won't even need to consider (6,5) and (7,4) ordered pairs as possible combinations of L,B since the prime factors 5 and 7 are not present in the prime factorisation of 144.

No need to consider 9,2 also as all sides are longer than 2cm.

No need to consider 10,1 also since 10 = 5×2, and 5 is not present in the prime factorisation as mentioned previously.

is it something like guess and check?

If you use his method, its guess and check partially.

If you use my method, it's elimination of options.

The low values and limited options for the guessing make the guess and check process a viable working method for this question.

Another way of doing using inequalities :

L + B = 11
L = 11 - B
B = 11 - L


Since B > 2 ,then 11 - B < 9
L < 9

But since L > 2 as well, then 2 < L < 9

3 ≤ L ≤ 8
(since the individual blocks are 1cm (whole number, the sides of the cuboids must have a whole number length)


Likwise, since L > 2, then 11 - L < 9
B < 9

Since B > 2, then 2 < B < 9
3 ≤ B ≤ 8


Next,

Since L > B, then 6 ≤ L ≤ 8
Which means 3 ≤ B ≤ 5


Now H = 144/LB

If we look at the range of L and B, then we can say that :

8 × 3 ≤ LB ≤ 6 × 5

24 ≤ LB ≤ 30

Since H = 144/LB,

144/30 ≤ 144/LB ≤ 144/24

4.8 ≤ H ≤ 6

Since H is a whole number, then 5 ≤ H ≤ 6

Since 144 does not contain prime factor 5, then H = 6

@Eric : I believe the question's setting intended for them to use prime factorisation to pick the right answer via logical elimination of prime factor options

As for partial guess and check method, not sure if the full 2 marks will be given.

LockB you can check with your teacher.

for the prime factorisation method, i dont really understand the part on combining some of the 2s... why is 1 of the "2" and "3" not combined with the other ones

Easy.

144 = 2 × 2 × 2 × 2 × 3 × 3

Remember that :

Volume , V = L × B × H

So, you want to express your volume as a product of 3 terms.

Since you need to express it as a product of 3 terms, and none of them can be 2, (because all sides are greater than 2cm so 2 is not a possible side length),

Then you'll need to 'combine' them.


Now, 2 × 2 × 2 × 2 × 3 × 3 can be rewritten in many ways :

2⁴ × 3 × 3 = 16 × 3 × 3

2² × 2² × 3² = 4 × 4 × 9

(2 × 3) × (2 × 3) × 2² = 6 × 6 × 4

And many more.


But, you must make sure that :

① L + B = 11
② both L and B comprise of the above prime factors.

The only combination that satisfies these conditions is :

L = 2³ = 8, B = 3

So,

144 = L × B × H

= 2 × 2 × 2 × 2 × 3 × 3

= 2³ × 3 × (2 × 3)

= 8 × 3 × 6

To answer your question :

Three '2's are combined/multiplied together to get 8.

One '3' and one '2' are combined to get 6.

The remaining 3 is the breadth.


We only need make sure that no '2's are left alone

i.e we don't want 144 = 2 × __ × __

Since 2 cannot be the length of any side as inferred from the question.

thx, i understand it now :)
by the way how do i integrate 8(x-1)/sqrt [(2x-1)^3]?the fraction is complicated and i cant seem to find any methods to integrate a complicated fraction...

Typically there is a previous part to this, something along the lines of perhaps x / sqrt (2x - 1)

oops i forgot about the previous part

btw for the above qn (the one i posted), why is the volume of the cuboid assumed to be 144cm^3 when they only said 144 1cm cubes?

Er... its obvious.

Liu has 144 one-centimetre cubes.
He arranges all of the cubes into a cuboid.

key phrase here is 'all of them'

Volume of each cube = 1cm × 1cm × 1cm
= 1cm³

Volume of all 144 cubes = 1cm³ × 144
= 144 cm³

Since he used all the cubes to form that cuboid, that cuboid's volume is equal to the total volume of the 144 cubes.

Which is 144 cm³

And since the cuboid is formed by those cubes, the cuboid's sides (length, breadth, height) have to be in multiples of 1 cm.

i.e whole numbers.

We can't have decimals or fractions.

hi, i've been trying to prove that f'(x) = 2e^x + e^-2x haa no stationary point got instead of getting the final as e^3x = - 2,im getting e^3x = - 1/2.what is wrong with my working?

at stationary points, f'(x) = 0
2e^x + e^-2x =0
2e^x + 1/e^2x = 0
2e^3x = - 1
e^3x = - 1/2 (rej)

on my second attempt, i got e^-3x = - 2 instead
2e^x + e^-2x = 0
e^-2x = -2e^x
e^-2x / e^x = - 2
e^-3x = - 2

i cant seem to get e^3x = - 2

What's the original question?

If you're asked to show f'(x) = 2eˣ + e⁻²ˣ has no stationary points, it means to show that df'(x) / dx ≠ 0 (also can be written as f''(x) ≠ 0), and not show f'(x) ≠ 0.

It would mean to find the stationary points of f'(x) , not f(x).

However, if you were asked to show f'(x) = 2eˣ + e⁻²ˣ ≠ 0 , then I think it's the answer key that is erroneous.

"the curve y=f(x) is such that f'(x) = 2e^x + e^-2x. explain why the cuve y=f(x) has no stationary points"
hence it is to use the given equation to show that it has no results but i cant get the correct working (as shown above)...

Then your answer is correct (either approach you use)

does that mean my answer of =-1/2 (re) is correct and not - 2??

Yes. Already mentioned that either approach you used is correct.

ohh tyvm :)

Alternatively,


f'(x) = 2eˣ + e⁻²ˣ


Now,

eˣ > 0 for all real x.

So, 2eˣ also > 0 for all real x.
(it's just multiplying by a positive number 2, which makes it still positive)


e⁻²ˣ = 1/e²ˣ = 1/(eˣ)²

The square of a positive value is still positive, so e²ˣ > 0

Taking the reciprocal of a positive value still gives a positive result,

so e⁻²ˣ = 1/e²ˣ > 0 for all real x



Then this implies that :

f'(x) = 2eˣ + e⁻²ˣ > 0 for all real x
(Adding two positive values still gives you a positive value)

So f'(x) ≠ 0 for all real x. (Can never be zero since it is always positive)

Hence, there are no stationary points of f(x)

i cant seem to get e^3x = - 2

The answer key seems to be flawed, and I have reason to believe that e^3x = -1/2.

But still, I would just use the fact that e^x > 0 for all real values of x (just as a^x > 0 for all real values of x and for any positive index a.

thx :) i wanted to do the method of showing that e^x/-x will forever be >0,but once i see the answer key i thought that method cant be used...

by the way why is ln(cos pi/4) = -1/2ln2?

This type of 'explain why' question can have different answers/explanation.

It's not those fixed answer ones .

As for your question,

cos π/4 = 1/√2 = 1 / 2¹/² = 2-¹/²

So ln(cos π/4) = ln 2-¹/²

= -½ ln 2