Preliminary mathematics marathon (1 Viewer)

[hscishard]

ehhh how you get R = root 1+3
i thought a = root 6 b = root 2
R = root (6+2)
i'm lost
link me to some tutorial site so i can learn how to deal with the (2x) part maybe

I divided root 2 on both sides first. Mkes it more simple

 

[super.muppy]

Explain carefully why | x + a | + | x + b | (for some a =/= b) is always strictly positive.
assume | x + a | > 0 | x + b | > 0 for some a =/= b
a =/= b =/=0 , as well as all real values,
| a | =/= | b |
since | x + a | > 0 , | a | =/= 0
therefore | x + a | > 0 is true
similarly, | x + b | > 0 is true
therefore, | x + a | + | x + b | (for some a =/= b) is always strictly positive.

it is still valid if a or b = 0 therefore line 2 of your working out is incorrect
|x +b| or |x+a| can be = 0 therefore line 1 of your working out is incorrect

 

[Trebla]

Explain carefully why | x + a | + | x + b | (for some a =/= b) is always strictly positive.
assume | x + a | > 0 | x + b | > 0 for some a =/= b
a =/= b =/=0 , as well as all real positive values,
| a | =/= | b |
since | x + a | > 0 , | a | =/= 0
therefore | x + a | > 0 is true
similarly, | x + b | > 0 is true
therefore, | x + a | + | x + b | (for some a =/= b) is always strictly positive.

Um well first of all, you've made some assumption and then concluded with those assumptions leading to a circular argument. Also, the values of a and b can be anything whether it be zero, positive or negative numbers.

I'm very tempted to give a hint but that will give away the solution...will just see what others suggest

 

[hscishard]

|x+a| >= 0
|x+b|>= 0
|x+a|+|x+b|>=0

But b =/=a
Therfore |x+a| =/= |x+b|
Then |x+a| + |x+b| =/= 0

Therefore from the 3rd line, |x+a|+|x+b|>0 i.e. is positive

Can that be right?

 

[fullonoob]

| x + a | + | x + b |
=sqrt (x+a)^2 + root (x+b)^2
= (x+a)^2 > 0 and (x+b)^2>0
x+a > 0 and x+b > 0
x> - a , x> -b
condition : a =/= b
since = (x+a)^2 > 0 and (x+b)^2>0
| x + a | + | x + b | > 0
if wrong clue me

 

[Trebla]

|x+a| >= 0
|x+b|>= 0
|x+a|+|x+b|>=0

But b =/=a
Therfore |x+a| =/= |x+b|
Then |x+a| + |x+b| =/= 0

Therefore from the 3rd line, |x+a|+|x+b|>0 i.e. is positive

Can that be right?

Very close! Can you provide a rigorous mathematical justification for that conclusion (based on the previous line) you made in bold? In other words explain carefully why | x + a | =/= | x + b | implies | x + a | + | x + b | =/= 0.

However, I can tell you now that it is possible to prove | x + a | + | x + b | =/= 0 easily without even needing to state | x + a | =/= | x + b |. It involves using a certain method of proof

Just another hint: it does involve an assumption but not the assumption that | x + a | and | x + b | are both positive...(and no it does not involve induction lol)

 

[Trebla]

| x + a | + | x + b |
=sqrt (x+a)^2 + root (x+b)^2
= (x+a)^2 > 0 and (x+b)^2>0
x+a > 0 and x+b > 0
x> - a , x> -b
condition : a =/= b
since = (x+a)^2 > 0 and (x+b)^2>0
| x + a | + | x + b | > 0
if wrong clue me

(x - a)² for example isn't necessarily always positive, it is non-negative. Also, when you solve inequalities with quadratics there are always two cases to consider (note that x + a < 0 for example is perfectly valid).

The first part of hscishard's proof is the starting point. So it goes:
| x + a | ≥ 0 and | x + b | ≥ 0
=> | x + a | + | x + b | ≥ 0

This means we have to prove that the equality cannot hold (i.e. strict inequality must hold) for a =/= b.

 

[hscishard]

Very close! Can you provide a rigorous mathematical justification for that conclusion (based on the previous line) you made in bold? In other words explain carefully why | x + a | =/= | x + b | implies | x + a | + | x + b | =/= 0.

However, I can tell you now that it is possible to prove | x + a | + | x + b | =/= 0 easily without even needing to state | x + a | =/= | x + b |. It involves using a certain method of proof

Let b=a+c

|x+a| + |x+(a+c)|>=0
If x+a =0, which will be|0|+|c|, which is greater than 0
If x+a+c=0, x+a=-c which will be |-c| + |0| which is positive

 

[Trebla]

Let b=a+c

|x+a| + |x+(a+c)|>=0
If x+a =0, which will be|0|+|c|, which is greater than 0
If x+a+c=0, x+a=-c which will be |-c| + |0| which is positive

Interesting approach, which is perfectly fine though it wasn't what I had in mind which was:

Suppose | x + a | + | x + b | = 0
=> | x + a | = - | x + b |
Since | x + a | ≥ 0 then - | x + b | ≥ 0 which is true only when x = - b which implies
| - b + a | = 0
=> a = b
But a =/= b so there is a contradiction, hence | x + a | + | x + b | =/= 0

Therefore | x + a | + | x + b | > 0

Anyway, I'll stop pretending to be a strict pure mathematician and give an applied mathematics question!

Suppose that there are two products A and B. A typical consumer that purchases these two products gains utility (which is a measure of "happiness") from them such that U = x₁x₂ where x₁ is the total quantity of product A purchased and x₂ is the total quantity of product B purchased. Suppose the price of product A is $1 and the price of product B is $2.

Suppose Michael is one of these typical consumers with such a utility function. Unfortunately, Michael only has $20 to purchase the two products which means he can only purchase a limited quantity of each product. Assume that he uses the entire $20 in his purchase.

(i) Write an expression in terms of x₁ and x₂ that represents the limited amount of each product he can purchase since he has only $20

(ii) Hence, find the quantities x₁ and x₂ such that his utility is maximised (i.e. find the most "happiness" Michael can attain with his limited $20 amount to spend)

 

[hscishard]

Never will have thought about that. Thx for the technique.
Wait is my thingy right?

 

[Trebla]

It is right, but it's not quite complete. You also need to show if the expressions x + a and x + a + c are non-zero then equality to zero cannot hold. Proof by contradiction is probably more efficient.

 

[jeshxcore]

Interesting approach, which is perfectly fine though it wasn't what I had in mind which was:

Suppose | x + a | + | x + b | = 0
=> | x + a | = - | x + b |
Since | x + a | ≥ 0 then - | x + b | ≥ 0 which is true only when x = - b which implies
| - b + a | = 0
=> a = b
But a =/= b so there is a contradiction, hence | x + a | + | x + b | =/= 0

Therefore | x + a | + | x + b | > 0

Anyway, I'll stop pretending to be a strict pure mathematician and give an applied mathematics question!

Suppose that there are two products A and B. A typical consumer that purchases these two products gains utility (which is a measure of "happiness") from them such that U = x₁x₂ where x₁ is the total quantity of product A purchased and x₂ is the total quantity of product B purchased. Suppose the price of product A is $1 and the price of product B is $2.

Suppose Michael is one of these typical consumers with such a utility function. Unfortunately, Michael only has $20 to purchase the two products which means he can only purchase a limited quantity of each product. Assume that he uses the entire $20 in his purchase.

(i) Write an expression in terms of x₁ and x₂ that represents the limited amount of each product he can purchase since he has only $20

(ii) Hence, find the quantities x₁ and x₂ such that his utility is maximised (i.e. find the most "happiness" Michael can attain with his limited $20 amount to spend)

Let the number of units of x1 purchased = z
Thus the number of x2 units purchased = (20 - z)/2

Therefore utility = z(x1) * ((20-z)/2)(x2)
du/dz = x1((20-z)/2)(x2) + z(x1) * (-1/2)(x2)
= (20.x1.x2)/2 - (z.x1.x2)/2 - (z.x1.x2)/2
= (20.x1.x2)/2 - z.x1.x2
= 0 at stationary points
therefore: 0 = (20.x1.x2)/2 - z.x1.x2
z.x1.x2 = (20.x1.x2)/2
2z.x1.x2 = 20.x1.x2
2z = 20
z = 10

SECOND DERIVATIVE:
du2/dz2 = -x1.x2 < 0 for all values of z
THEREFORE z = 10 is a MAXIMUM

therefore the amount of x1 bought = 10 units
the amount of x2 bought = 5 units


i hope that shit is right lol

 

[jeshxcore]

Let the number of units of x1 purchased = z
Thus the number of x2 units purchased = (20 - z)/2

Therefore utility = z(x1) * ((20-z)/2)(x2)
du/dz = x1((20-z)/2)(x2) + z(x1) * (-1/2)(x2)
= (20.x1.x2)/2 - (z.x1.x2)/2 - (z.x1.x2)/2
= (20.x1.x2)/2 - z.x1.x2
= 0 at stationary points
therefore: 0 = (20.x1.x2)/2 - z.x1.x2
z.x1.x2 = (20.x1.x2)/2
2z.x1.x2 = 20.x1.x2
2z = 20
z = 10

SECOND DERIVATIVE:
du2/dz2 = -x1.x2 < 0 for all values of z
THEREFORE z = 10 is a MAXIMUM

therefore the amount of x1 bought = 10 units
the amount of x2 bought = 5 units


i hope that shit is right lol

actually perhaps cos it's a parabola, cut that shit out

the parabola is negative, and hence has a maximum

the maximum occurrs half way between the 2 "z" intercepts


z(x1) * ((20-z)/2)(x2) = 0
z = 0, 20
therefore: z that maximises U = (20 - 0)/2
= 10

 

[ohexploitable]

(i) Write an expression in terms of x₁ and x₂ that represents the limited amount of each product he can purchase since he has only $20

amount of A + amount of B = $20

amount of $1 + amount of $2 = $20

x₁*$1 + x₂*$2 = $20

x₁ + 2x₂ = 20 ---- (1)

(ii) Hence, find the quantities x₁ and x₂ such that his utility is maximised (i.e. find the most "happiness" Michael can attain with his limited $20 amount to spend)

let x₂ = x

U = x₁x ---- (2)

from (1): x₁ = 20 - 2x ---- (1a)

subs. (1a) into (2):

U = x(20 - 2x)

U = -2x²+20x ---- (2a)

sad face parabola, maximum at dU/dx = -4x + 20 = 0

maximum at x = 5 ---- (3)

subs. (3) into (1a)

x₁ = 10

.: x₁ = 10, x₂ = 5

 

[ohexploitable]

therefore the amount of x1 bought = 10 units
the amount of x2 bought = 5 units

Yay! I got the same.

 

[edmundsung]

another question

 

[ohexploitable]

Re: another question

AP || BQ (co-interior angles are supplementary)

Construct BC ⊥ AP

BC = 15 cm (Pythagorean theorem)

.: PQ = 15 cm (opposite sides of a rectangle are equal)

fixed retardedness

 

[edmundsung]

Re: another question

AP || BQ (co-interior angles are supplementary)

Construct BC ⊥ AP

BC = √777 cm (Pythagorean theorem)

.: PQ = √777 cm (opposite sides of a rectangle are equal)

just wondering how u got BC=√777 cm by pyth. theorem.
BC=√777 =27.87(cor. to 2 dp)
which is larger than AB (17cm) lol considering AB is the hypotenuse @@

 

[ohexploitable]

Re: another question

just wondering how u got BC=√777 cm by pyth. theorem.
BC=√777 =27.87(cor. to 2 dp)
which is larger than AB (17cm) lol considering AB is the hypotenuse @@

lol I read "distance between the centres" as "distance between the circles", haha, I thought AB was 17+10+2=29. lol.

So yeah, PQ is 15 cm.

 

[jeshxcore]

Re: another question

AP || BQ (co-interior angles are supplementary)

Construct BC ⊥ AP

BC = √777 cm (Pythagorean theorem)

.: PQ = √777 cm (opposite sides of a rectangle are equal)

isnt BC = √225 = 15 cm...


EDIT: oh woops already sorted out LOL