Mathematics Question Thread

[katnisschung]

hey it won't let me attach my working so i'll try explain it best below
also attached the original q

2x+2y+4x=10
y=5-3x

let a=area

a= (5-3x) (x) + (x)^2
a=5x-2x^2

derivative of a=0 at stationary points

5-4x=0 therefore, x=1.25

used 2nd derivative to test nature
max as -4

sub x=1.25 into original equation

thus y=1.25
soz for the dodgy quality

[jamonwindeyer]

hey it won't let me attach my working so i'll try explain it best below
also attached the original q

2x+2y+4x=10
y=5-3x

let a=area

a= (5-3x) (x) + (x)^2
a=5x-2x^2

derivative of a=0 at stationary points

5-4x=0 therefore, x=1.25

used 2nd derivative to test nature
max as -4

sub x=1.25 into original equation

thus y=1.25
soz for the dodgy quality

Cheers! You are correct - There is nothing prohibiting \(y=x\) - A square is still a rectangle

[katnisschung]

aha yeah i thought that too thinking they were out to trick me..

[katnisschung]

hi hi im stuck on this q

a rectangle is cut from a circular disc of 6cm
find the area of the largest rectangle that can be
produced

so i did

x^2+y^2=36
a=xy

rearranged the first to find y in terms of x for area

a=x(36-x^2)^1/2
derivate of a=0 at stationary points

found the derivative but having trouble finding x

could somebody pls show the full working out...terrible at algebra

this was as far as i got

(36-x^2)1/2 - (x^2)/(36x-x^2)^1/2 = 0

Thanks!

[jakesilove]

hi hi im stuck on this q

a rectangle is cut from a circular disc of 6cm
find the area of the largest rectangle that can be
produced

so i did

x^2+y^2=36
a=xy

rearranged the first to find y in terms of x for area

a=x(36-x^2)^1/2
derivate of a=0 at stationary points

found the derivative but having trouble finding x

could somebody pls show the full working out...terrible at algebra

this was as far as i got

(36-x^2)1/2 - (x^2)/(36x-x^2)^1/2 = 0

Thanks!

So, we have



So,




If we assume that x cannot equal six (which is clearly can't, else the denominator would equal zero) then we can just multiply this out




Then, you can use the quadratic formula etc. to solve for x

[katnisschung]

thanks jake!

stuck on another 

a sufboard is in the shape of a rectangle and semi circle
(half a circle on top of a rectangle). The perimeter is to be 4m
find the max area of the surboard correct to 2 dp

im really having trouble differentiating this
quick run through of my working out

p=2l+b+1/2 2 b
found b as
b=4-2l/1+

substituted to find A
attached image

[katnisschung]

So, we have



So,




If we assume that x cannot equal six (which is clearly can't, else the denominator would equal zero) then we can just multiply this out




Then, you can use the quadratic formula etc. to solve for x

sorry what did u multiply out??

[jakesilove]

thanks jake!

stuck on another 

a sufboard is in the shape of a rectangle and semi circle
(half a circle on top of a rectangle). The perimeter is to be 4m
find the max area of the surboard correct to 2 dp

im really having trouble differentiating this
quick run through of my working out

p=2l+b+1/2 2 b
found b as
b=4-2l/1+

substituted to find A
attached image

It sounds like you need to go back and revise differentiation. We have the function



So,



by the chain rule for the second term. Then, set this equal to zero, and solve for l.

For the last question, I multiplied everything by the denominator.

[Sukakadonkadonk]

Hey guys

Could someone please explain to me why:

(2/3)(x)^3/2

equals to:

2(x)^3

??

[RuiAce]

Hey guys

Could someone please explain to me why:

(2/3)(x)^3/2

equals to:

2(x)^3

??

It doesn't

Where did you get this from

[Mahan]

Hi, is there anyway to solve this question without drawing the actual cos graph and y=+- 1?

Well, it addition to using , there is another algebraic way to solve this:
using double eagle formula.

[Sukakadonkadonk]

It doesn't

Where did you get this from

Ooops sorry, typo. I meant:

(2/3)(x)^3/2

Equals to:

2(x)^1/2

[RuiAce]

Ooops sorry, typo. I meant:

(2/3)(x)^3/2

Equals to:

2(x)^1/2

They still don't equal.

[Sukakadonkadonk]

They still don't equal.

Ok so is this incorrect working then?
I've seen this twice already in two papers so I'm really confused.

The answer is boxed.

Thanks.

[RuiAce]

Ok so is this incorrect working then?
I've seen this twice already in two papers so I'm really confused.

The answer is boxed.

Thanks.

Line 3 is wrong. It should not be there.