Geom. Applications of Calc. (2 Viewers)

[-pari-]

i seem to have hit a point where i'm getting all the questions wrong, so i'm obviously doing something wrong but i can't figure out what...

1) A farmer wants to make a rectangular paddock with an area of 4000m^2. She wants a minimum perimeter.
a) we figure out the equation of the perimetre iss.....

P= 2x + 8000
x

b) [The part i get stuck at] Find the dimensions that will give the minimum perimetre (1dp) [Answer: 63.2m by 63.2m)



2) the forumla for surface area of a cylinder is given by SA = 2(pi)r (r + h)
the volume = 54 (pi) metres^3

surface area = 2(pi)r^2 + 108(pi)
r

Find the radius that gives the minimum surface area.
[Answer: Radius = 3m]


11) A cylinder is requried to hold 8600m^3 of wheat.

equation for the surface area in terms of the base and the radius :
SA = 2(pi)r^2 + 17200
r

Find the minimum surfarce area required to hold this amount of wheat (nearest square metre)

 

[jyu]

i seem to have hit a point where i'm getting all the questions wrong, so i'm obviously doing something wrong but i can't figure out what...

1) A farmer wants to make a rectangular paddock with an area of 4000m^2. She wants a minimum perimeter.
a) we figure out the equation of the perimetre iss.....

P= 2x + 8000
x

b) [The part i get stuck at] Find the dimensions that will give the minimum perimetre (1dp) [Answer: 63.2m by 63.2m)



2) the forumla for surface area of a cylinder is given by SA = 2(pi)r (r + h)
the volume = 54 (pi) metres^3

surface area = 2(pi)r^2 + 108(pi)
r

Find the radius that gives the minimum surface area.
[Answer: Radius = 3m]


11) A cylinder is requried to hold 8600m^3 of wheat.

equation for the surface area in terms of the base and the radius :
SA = 2(pi)r^2 + 17200
r

Find the minimum surfarce area required to hold this amount of wheat (nearest square metre)

Applications of differentiation

1) dP/dx = 2 - 8000/x^2 = 0 etc

2) dA/dr = 4(pi)r - 108(pi)/r^2 = 0 etc

11) Similar to 2), find r and then the minimum A.

 

[Riviet]

Note that when asked to find the maximum or minimum with these kind of problems, you will nearly always need to differentiate your equation and let it equal to zero as jyu has shown.

Strictly speaking, you should also prove that it's a max/min by using either the first derivative or second derivative test.

 

[-pari-]

yeah i did that....
i realised i was differentiating wrong.

this one stumped me as well:
]

A surfboard is in the shape of a rectangle with a semicircle stuck on on of its (shorter) sides.. (you get the picture...)
the perimetre is 4m. find the maximum area of the surfboard (2dp)
----
this is what i've done:

P = pr + 2x (the lengths) + y (the shorter side)
pr + 2x + y = 4.

y = 4 - 2x - pr

Area = pr^2 + xy
2

sub in y

A = pr^2 + x (4 - 2x - pr)
2

= pr^2 + 4x - 2x^2 - prx
2

& then you differentiate it and whatnot, but my problem is how do i get the first derivative to equal zero, and then extract a value for x ??
my x value always ends up with pr in it...which isn't much good....

 

[-pari-]

k well the fractions didn't turn out right but all the little 2's are meant to be under the pr^2

 

[jyu]

yeah i did that....
i realised i was differentiating wrong.

this one stumped me as well:
]

A surfboard is in the shape of a rectangle with a semicircle stuck on on of its (shorter) sides.. (you get the picture...)
the perimetre is 4m. find the maximum area of the surfboard (2dp)
----
this is what i've done:

P = pr + 2x (the lengths) + y (the shorter side)
pr + 2x + y = 4.

y = 4 - 2x - pr

Area = pr^2 + xy
2

sub in y

A = pr^2 + x (4 - 2x - pr)
2

= pr^2 + 4x - 2x^2 - prx
2

& then you differentiate it and whatnot, but my problem is how do i get the first derivative to equal zero, and then extract a value for x ??
my x value always ends up with pr in it...which isn't much good....

Don't forget y = 2r, P = pr + 2x + 2r =4, r = (4 -2x)/(p + 2),

A = 0.5pr^2 + 2rx = etc

 

[imprsiv]

Those max and min problems are a biatch did them last year and could never consistenly get them right. Can do them alright now after being taught them a second time. (dropped from MX1 to 2 unit) I found if you keep practicing you would eventually work out how to do them.

I'm sure this is how you do it.
1) Express the quantity to be maximsed or minimised in terms of one variable
2) Differentiate with respect to the variable, make this function = 0, then solve
3) Test to show max or min occurs in 2nd derivative{ f``(x)}. Show change of sign using 1st derivative.

Question to show my steps
1)The council wanted to make a rectangular swimming area at a beach using a straight cliff on one side and a length of 300m of shark proof netting. For the either 3 sides, what are the dimensions of the rectangle that enclose the greatest(maximum area)

1) step one create a variable

p= 2y+x=300
x=300-2y
Area=xy

But x=300-2y
A=(300-2y)y
A= 300y-2y^2


2) differentiate

f(x)= 300y-2y^2
f`(x)= 300-4y
make f`(x)=0
0=300-4y
y=75

3) f``(x) = -4

Therefore y=75 (max value) {greatest}

 

[-pari-]

^^ thanks imprsv

Don't forget y = 2r,

genius. i didn't notice that! thanks

 

[-pari-]

question!
there's a right angled triangle with hypotenuse 5m,
A = 1/2.x.(25-x^2)^1/2
1) find the greatest possible area of the triangle.
[answer: 6.25m^2)
i know what to do...as in differentiate etc, but then i dont seem to be doing it right for some reason. i have a feelin my differentiation is wrong ...

2) find the first and second derviatives of

5-x
(4x^2 + 1)^3

i keep getting (for the first derivative):

-4x^2 - 61 + 12x
(4x^2 +1) ^4

but the answer is:

20x^2 - 120x - 1
(4x^2 + 1)^4

....???

 

[Riviet]

1) Make sure you have used the product rule for x/2 and (25-x²)^1/2 and don't forget to use the chain rule for (25-x²)^1/2.

2) Let y=(5-x)/(4x²+1)³

Then dy/dx = [-(4x+1)³(1) - 3(5-x)(4x²+1)².8x] / [(4x²+1)³]²

= {(4x²+1)²[-(4x+1) - 24x(5-x)]} / (4x²+1)⁶

= (-4x² - 1 - 120x + 24x²) / (4x²+1)⁴

= (20x² - 120x - 1) / (4x²+1)⁴ #

Hope that helps

 

[-pari-]

k, with that question (1), i'm using the chain rule and the product rule and tis not working. help?

i get

A' = -x + 2500 - 4x^4

and i need to make that = 0 to find the greatest value of x.

so i'm stuck...is my differentiation wrong or...?

 

[Riviet]

For Q1, are you sure your equation is correct?

Because dA/dx=1/2.{-2x/2.(25-x²)^-1/2 + (25-x²)^1/2} = 0

This simplifies to x²+x-25=0, which has irrational roots.

 

[jyu]

For Q1, are you sure your equation is correct?

Because dA/dx=1/2.{-2x/2.(25-x²)^-1/2 + (25-x²)^1/2} = 0

This simplifies to x²+x-25=0, which has irrational roots.

The equation is correct.

dA/dx = 1/2[-x^2 /sqrt(25 - x^2) + sqrt(25 - x^2)] = 0

x^2 = 25/2

so max area = 1/2 (5/sqrt2)(5/sqrt2) = 6.25

Happy 2007 :wave:

 

[-pari-]

(1) Find the exact value of f"(2) if f(x) = x.rt(3x-4)

i keep getting 34/rt2 = 17rt2

but its wrong.

answer: 3rt2/8

(2) Find the turning point of the curve y = 3x^4 + 1 & determine its nature.
i know how to do it....BUT!

you find y'
let y'= 0 for turning points.
in this case, x = 0

so when when x = 0, y" = 0

so when y" = 0 it should be a point of inflexion if concavity changes on both sides of the point.

BUT concavity doesn't change. on both LHS and RHS y">0
so yea its a minima point,

but isn't that contradictory?
that y" = 0.....and yet on either side of the point, the concavity remains the same.
i dont get how that works.....

(3) f(x) = 2x^5 + 3
is there a turning point and if so determine its nature.

i keep gettin
y' = 10x^4 + 3
which, when u equate it to zero...x = not defined....?

but answer is : point of inflexion at (0,3)

(4) Find any turning points on the curve y = (4x^2 - 1)^4
i've redone this like five times. keep getting the same wrong answer.

answers: (0, 1) maximum point [i got this one]
(0.5, 0) min point
(-0.5, 0) min point.

(5) also...is there any difference between a 'horizontal' point of inflexion and just a plain point of inflexion?
----

 

[bos1234]

what does point of inflexion mean again?

If y''< o what does that say about point of inflexion?

 

[pLuvia]

1) I got what they wanted, it's probably just an algebraic mistake you have somewhere in your working

2) This is a horizontal point of inflection

3) and 4) Just probably algebraic mess ups, just check over it

5) A horizontal point of inflexion you can see it as a horizontal line on the graph like this
\_ <-- This underscore is the HPOI
...\

A plain POI is just just a "continuous" line without the bump in between, it's when the concavity is changed

 

[bos1234]

(1) Find the exact value of f"(2) if f(x) = x.rt(3x-4)

i keep getting 34/rt2 = 17rt2

but its wrong.

answer: 3rt2/8

(2) Find the turning point of the curve y = 3x^4 + 1 & determine its nature.
i know how to do it....BUT!

you find y'
let y'= 0 for turning points.
in this case, x = 0

so when when x = 0, y" = 0

so when y" = 0 it should be a point of inflexion if concavity changes on both sides of the point.

BUT concavity doesn't change. on both LHS and RHS y">0
so yea its a minima point,

but isn't that contradictory?
that y" = 0.....and yet on either side of the point, the concavity remains the same.
i dont get how that works.....

(3) f(x) = 2x^5 + 3
is there a turning point and if so determine its nature.

i keep gettin
y' = 10x^4 + 3
which, when u equate it to zero...x = not defined....?

but answer is : point of inflexion at (0,3)

(4) Find any turning points on the curve y = (4x^2 - 1)^4
i've redone this like five times. keep getting the same wrong answer.

answers: (0, 1) maximum point [i got this one]
(0.5, 0) min point
(-0.5, 0) min point.

(5) also...is there any difference between a 'horizontal' point of inflexion and just a plain point of inflexion?
----

(3) You cant differentiate the constant...

 

[pLuvia]

(3) You cant differentiate the constant...

When you differentiate a constand it becomes zero!

 

[-pari-]

^ o yeaa.. ahaha..oops

2) This is a horizontal point of inflection

answer says minima?

 

[bos1234]

(2) Find the turning point of the curve y = 3x^4 + 1 & determine its nature.
i know how to do it....BUT!

you find y'
let y'= 0 for turning points.
in this case, x = 0

so when when x = 0, y" = 0

so when y" = 0 it should be a point of inflexion if concavity changes on both sides of the point.

BUT concavity doesn't change. on both LHS and RHS y">0
so yea its a minima point,

but isn't that contradictory?
that y" = 0.....and yet on either side of the point, the concavity remains the same.
i dont get how that works.....

(3) f(x) = 2x^5 + 3
is there a turning point and if so determine its nature.

i keep gettin
y' = 10x^4 + 3
which, when u equate it to zero...x = not defined....?

but answer is : point of inflexion at (0,3)


(5) also...is there any difference between a 'horizontal' point of inflexion and just a plain point of inflexion?
----

Point of inflexion means that the concavity changes!

For example, a cubic curve. Its composed of a concave up curve and then slowdly turns into a concave down curvve and some point (x.y)

So (2) Find the turning point of the curve y = 3x^4 + 1 & determine its nature.

So y'=12x^3

y''=36x^2

x -1 0 1
y'' >0 0 > 0

when y'' > 0 it means its concave. So it never becomes concave down...

y"< 0 means its concave down


--------

Horizontal point of inflexion means its a turning point and also it changes concavity at that point