The water takes the shape of a cone as well. This cone shape is geometrically similar to the conical container.

The ratio of their radii = ratio of their height

Let radius of water's cone shape be r.

So r : 10 = h : 30

3r : 30 = h : 30

Therefore, 3r = h
r = ⅓h

Or

r/10 = h/30
30r = 10h
r = ⅓h


Volume of water = ⅓πr²h
= ⅓π(⅓h)²h
= ⅓π(1/9 h²)h
= 1/27 πh³

(Shown)

Volume of water, V = 1/27 πh³
(found earlier)

dV/dh = 3(1/27) πh²
= 1/9 πh² or πh²/9


dV/dt = 10cm³s-¹ (given)


dh/dt = dV/dt × dh/dV

= dV/dt ÷ dV/dh

= 10 cm³s-¹ ÷ πh² / 9

= 10 cm³s-¹ x 9/πh²

= 90cm³s-¹ / πh²


When r = 2cm, and having found that r = ⅓h,

2cm = ⅓h
h = 6cm


So, dh/dt = 90cm³s-¹ / πh²

= 90cm³s-¹ / π(6cm)²

= 90cm³s-¹ / 36πcm²

= 5/2π cm s-¹

PQR is an isosceles triangle since PQ = PR

Firstly , divide PQR into 2 identical right-angled triangles by drawing the perpendicular line from P to the midpoint of QR. Call the midpoint M.

So QM = RM and QR = 2QM = 2 RM

QM = QR ÷ 2
= (16 - 6x) ÷ 2
= 8 - 3x


Next, use Pythagoras' theorem to find PM, which is the height of PQR.


QM² + PM² = PQ²

(8 - 3x)² + PM² = (5√(x²+4))²

(8 - 3x)² + PM² = 25(x² + 4)

64 - 48x + 9x² + PM² = 25x² + 100

PM² = 16x² + 48x + 36

PM² = (4x + 6)²

PM = 4x + 6


Area of triangle = ½ x base x height

= ½ x QR x PM

= ½ (16 - 6x)(4x + 6)

= (8 - 3x)(2)(2x + 3)

= 2(2x + 3)(8 - 3x)

(Shown)

Lastly,

A = 2(2x + 3)(8 - 3x)

A = 2(16x - 6x² + 24 - 9x)

= 2(24 + 7x - 6x²)

= 48 + 14x - 12x²

dA/dx = 14 - 24x


When dA/dx = 0,

14 - 24x = 0

24x = 14
x = 14/24 = 7/12

Second derivative test :

d²A/dx² = -24 < 0 (maximum)

When x = 7/12,

Maximum Area
= 2 (2(7/12) + 3) (8 - 3(7/12))
= 2 (50/12) (75/12)
= 7500/144

= 50 1/12 units

Edit : should be 50 1/12 cm² instead of units

For the second one, why is dV/dt x dh/dV = dV/dt ÷ dV/dh

And I don’t need to differentiate 10cm^3s^-1 ?

For the last one, why is dA/dX = 0 ?

Think of it like fractions. Numerator on one cancels out denominator on the other.

dh/dt = dV/dt x dh/dV

So the two dV will cancel out.

Now ,

dh/dV = 1 / dV/dh
(you should have already learnt why this is the case at O levels)


So ,

dh/dt = dV/dt x dh/dV

= dV/dt x 1 / dV/dh

= dV/dt ÷ dV/dh


Its similar to fractions multiplication. When you change the sign to division , you invert the numerator and denominator. Vice versa


Eg.

2/3 x 5/4 = 2/3 ÷ 4/5

1/6 ÷ 2/7 = 1/6 x 7/2

It is given that water is poured into the cone at the rate of 10cm³s-¹.


This is already implying that the volume of water in the cone will increase at 10cm³s-¹ (or 10cm³ per second) since no water is flowing out elsewhere.


Therefore the rate of change of volume with respect to time (dV/dt)

= 10cm³s-¹

There is no need to differentiate.

A maximum or minimum point occurs when the slope/gradient is 0 and is the point where there is a change in sign of the slope of gradient.

They are a subset of stationary points

(the other in the category being point of inflection/inflexion)


Since the question asks for maximum area, you'll need to differentiate A with respect to x first.


And when dA/dx = 0, it is where the area
does not increase anymore / does not decrease anymore.


The rate of change of area with respect to change in x is 0.


So solve for dA/dx = 0 to find the value of x where this occurs.

Then use thay value of x to find the area where dA/dx = 0

For this question it is at a maximum point (so maximum area)




An analogy for a maximum ppoint is the highest point of a peak, and for minimum point it is the lowest point of a valley.

Thanks for the explanation:)

Welcome.


An example to see how dV/dh = 1/ dh/dV :

Let's say V = 2h

dV/dh = 2


But if we do it the other way,

h = ½V

dh/dV = ½


2 and ½ are reciprocals of each other.
(1 ÷ 2 = ½
1 / ½ = 1 x 2/1 = 2)

When dV/dh = 2, the rate of change of V is twice the rate of change of h (increases/decreases twice as fast)


So this means the rate of change of h is half that the rate of change of V
(increases/decreases half as fast)

The changes are relative to each other

Got it