Math & Logic Challenge

The perimeter and diagonal length of a rectangle are respectively 100 units and x units. Find, in terms of x,  the area of the rectangle.  

Let us develop two ideas ‘simultaneously’.

Perimeter and area	Pythagoras
P = 2 L + 2 B = 100	L2 + B2 = x2
B = L – 50	L2 + (L – 50)2 = x2
A = L × B	L2 + L2 – 100L + 2500 = x2
A = L ( L – 50)
A = L2 – 50L	2L2– 100L = x2 – 2500

Consequently:

A = ½ (2L²– 100L)

A = ½ ( x² – 2500)  

 

This solution only works because of a fortuitous coincidence in the coefficients.

In general, we should ‘complete the square.’

2L²– 100L = x² – 2500

2(L²– 50L) = x² – 2500

2(L²– 50L + 0) = x² – 2500

2(L²– 50L + 625 - 625) = x² – 2500

2((L – 25)²– 625) = x² – 2500

2(L – 25)²– 2 × 625 = x² – 2500

2(L – 25)²– 1250 = x² – 2500

2(L – 25)² = x² – 2500 + 1250

2(L – 25)² = x² – 1250

 (L – 25)² = ½ x² – 625

L – 25 = √ (½ x² – 625)

L = 25 + √ (½ x² – 625)

Having obtained L, obviously we can calculate Lⁿ and kL, for various n and k.  However simplification would be tedious. 

In conclusion, competition maths is very different from exam maths. In a competition we are looking for creativity and flexibility.  In an exam we want regularity and conformity.

 

Y10 Add Maths Related Rates Quiz

Johhny asked Mary, why did Sir say: "You cannot step twice into the same river." I looked it up, replied Mary. He was quoting Heraclitus of Ephesus. It means that it is not the same river because the water you stepped into has already moved downstream and other water with possibly a different pH, temperature and biological oxygen demand has replaced it. Actually, said Johnny I think it means more than that. It is not the same “you” either. You have had the new experience of stepping into the polluted river, the chocolate bar you have eaten has released phenyl ethyl amine into your blood stream and an enzyme in one of trillions of your cell nuclei has been busy repairing a piece of DNA which was damaged by a cosmic ray that had travelled 15 billion light years from a dying star at the edge of the known universe!

“Somwhow you forgot to say I am also prettier than I was yesterday. What do you know about Parmenides?” asked Mary? “Is he the new student all the girls are crazy about?” replied Johnny. Of course not said Mary. He’s another Greek philosopher. In his poem, On Nature, he explains how reality is one and change is impossible. Our senses cannot perceive true reality. As we focus on different parts of the same one reality we just have the illusion that things have changed. For example, your iPod has all the songs stored in it at once, but you can only hear one chord at a time. In the same way, a superior being outside of time could perceive our wholes life at once.

In the debate between these philosophers from very different viewpoints, mathematics takes the middle ground. Change exists but behind change we a permanent form.

The area of a triangle (see diagram) is growing at 3 cm² per second.

i.e., dA/dt = 3 and the triangle remains self-similar

i) Find the rate of increase of the base, when the base is exactly 12 cm and the height is exactly 6 cm. [3]      

Note that h = ½ b  (Not just at time a particular time t but for the duration of growth of the triangle)

A = ½ b h

A = ½ b ( ½ b)

A = ¼ b²

dA/db = ½ b

dA/dt = dA/db × db/dt

3 = ½ (12) × db/dt

db/dt = ½

ii) Find the rate of increase of the height, when the base is exactly 12 cm and the height is exactly 6 cm.    [2]

Note that b = 2h and proceed as before

A = ½ b h

A = ½ (2h) ( h)

A = ½ (2 )h²

A = h²

dA/dh = 2h

dA/dt = dA/dh × dh/dt

3 = 2 (6) × db/dt

db/dt = ¼

Y10 Add Maths Max/min Quiz

A cuboid has a total surface area of 120 cm². Its base measures x cm by 2x cm and its height is h cm.

i) Obtain an expression for h in terms of x.                                                                                                     [2]

Make a sketch:

The area of the base is x × 2x = 2x².  The top face is the same.

The area of the right side is 2x × h = 2xh. The left face is the same.

The area of the front face is x × h = xh.  The back is the same.

Surface area = 120 = 4x² + 6xh

120 – 4x² = 6xh

h = (120 – 4x²)/6x

h = 20x^-1 – 2x/3

Given that the volume of the cuboid is Vcm³,

ii) show that V = 40x – 4 x³/3 .                                                                                                               [1]

V = x × 2x × h

V = 2 x² (20x^-1 – 2x/3)

V = 40x - 4 x³/3

Given that x can vary,

iii) show that V has a stationary value when h = 4x/3 .                                                                            [4]

dV/dx = 40 - 4x²

0 = 40 - 4x²

x = +√10  (reject the negative root)

Substitute this value of x into h = 20x^-1 – 2x/3

h = 20/√10 - 2√10/3 = 2 √10√10/√10 - 2 √10/3    (Since 20 = 2 × 10 and 10 itself = √10²)

h = 2√10 - 2√10/3

h = 6√10/3 - 2√10/3

h = 4√10/3

h = 4x/3      (Recall that x = √10).

iv) Find the stationary value of V.                                                                                                       [1]

V = 40x - 4 x³/3

= 40√10 - 4 √10 ³/3               ( √10) ³  = √10√10√10 = 10√10 )

= 40(√10 - √10)/3)

= 80√10/3

iv) Determine the nature of the stationary value. Show working.                                                          [2]

d²V/dx² = -8x = (negative)(positive) = (negative).    Note that x itself must be > 0, i.e., positive

The function is concave down. The turning point is a maximum.

A Max-min Quiz

Given V  = V_cylinder + V_hemisphere = 2880π 

a) Make h the subject of the equation:

2880π =  πr²h + (2/3) πr³  

Divide both sides by π

2880 = r²h + (2/3) r³  

h = {2880 – (2/3) r³}/ r² 

h = 2880/r²  – (⅔)r = 2880r^–2 – (⅔)r

b) Find the surface area as a function of r only.

Surface area, S = A_circle + A_rectangle + A_hemisphere

S = πr² + 2πrh + 2πr²

S = 3πr² + 2πrh

S = 3πr² + 2πr{2880r^–2 – (⅔)r}

S = 3πr² + 5760πr^–1 – (4π/3) r²

S = (5π/3) r² + 5760πr^–1                                                                           as required

c) Find the derivative of the surface area, the stationary point of the surface area function and the surfasce area at the stationary point.

dS/dr = 2(5π/3) r¹ + (-1)5760πr^–2 

dS/dr = (10 π /3) r – 5760πr^–2 

dS/dr = (10 π /3) r – 5760πr^–2 

find a common denominator: dS/dr = (10 π r³/3r²)– (17280π/3r²)

For a stationary point dS/dr = 0

For a fraction to equal 0, the numerator equals 0.

Therefore 0 = 10 π r³ – 17280π

17280 π = 10 π r³

r³  = 1728

r = 12

When r = 12, S = (5π/3) (12)² + 5760π/12 

S = 240 π + 480 π

S = 760 π   (~ 2260 but leave in terms of π)

A sketch:

From the graph the stationary point appears to be a minimum. To confirm this, find the second derivative:

d²S/dr² = 10 /3 + 11520 /r3

d²S/dr² = a  positive term + another positive term

Hence the curve is concave up.  It is confirmed that the stationary point is a minimum.

How to Use the Discriminant

Solve for k, given Y_Parabola = 1x² + (2k + 10)x + k² + 5,  tangent to the x-axis.

 Mary says “Obviously”, b² – 4ac = 0              (Johnny says, "This may not be obvious.  I will provide a note below.)

(2k + 10)2 – 4(1)( k2 + 5) = 0	Substitution of  a = 1, b = 2k + 10 and c = k2 + 5
4k2 + 40k + 100 –  4k2  – 20 = 0	Expand
40k = -80
k = -2	Done.

               

Note:  At the point where a line is tangent to a parabola, Y_Parabola = Y_Line. Therefore Y_Parabola – Y_Line = 0_.  But in this case the tangent line is the x-axis (given) so the line is simply y = 0.  (or, y = 0x + 0 if you want to be precise!) Since the gradient of y = 0 must be m = 0, we know the tangent is horizontal. This only happens at a stationary point.  But a parabola only has one stationary point (the maximum or minimum).  In this case the stationary point touches the x-axis so it is also the only root of the parabola.  Consequently the discriminant, b² – 4ac is 0.  (Recall that, if there are two real roots b² – 4ac > 0, if there are no real roots b² – 4ac < 0.)

Calculus has only been used in the background to understand the problem.   We never found the first derivative.  This suggests an alternative but more complicated approach.

Alternate method:

m = dy/dx = 2x + 2k + 10 

As noted above, m = 0  (The tangent is horizontal).

0 = 2x + 2k = 10

0 = 2(x + k + 5)

0/2 = 0 = x + k + 5

x =  – k – 5 

Substitute this x into Y_Parabola = 0             (The tangent is the x-axis which is the line y = 0).

1x² + (2k + 10)x + k² + 5 = 0

(–k – 5)² + (2k + 10) (–k – 5) + k² + 5 = 0

k² + 10k + 25 – 2k² – 10k –10k – 50 + k² + 5 = 0

20 = -10k

k = -2          As before. Undoubtedly there exist even more complicated solutions but Johnny will stop here.