Maxima + Minima

Makro · Mar 1, 2009

I may add to this thread.

So I struggle with these questions, I don't do too bad at the manipulating equation bit, but I just fail really badly on the calculus bit.

Anywho, I've got these two:

5. The height in metres of an object thrown up into the air is given by h = 20t - 2t² where t is time in seconds. Find the max height that the object reaches.

11. A soft drink manufacturer wants to minimise the amount of aluminium in its cans while still holding 375 mL of softdrink. Given that 375 mL has a volume of 375 cm³,
a) Show that the SA of a can is given by S = 2(pi)r² + 750/r
b) Find the radius of the can that gives the min SA

Thanks in advance.

EDIT: More :/

13. A rectangular prism with a square base is to have a surface area of 250 cm².
a) Show that the volume is given by V = (125x - x³)/2
b) Find the dimensions that will give the max volume.

lyounamu · Mar 1, 2009

5. The height in metres of an object thrown up into the air is given by h = 20t - 2t2 where t is time in seconds. Find the max height that the object reaches.

dh/dt = 20 - 4t
t = 5 when dh/dt = 0 (i.e. to find a stationary point)

second derivative = -4
meaning that t=5 is a maximum point

sub t= 5

h = 20 x 5 - 2.5^2 = 50

13. A rectangular prism with a square base is to have a surface area of 250 cm2.
a) Show that the volume is given by V = (125x - x3)/2
b) Find the dimensions that will give the max volume.

let height be y and base be x.

so 2x^2 + 4xy = 250
4xy = 250 - 2x^2
y = (250-2x^2)/4x

V = x^2 . y = (125x - x^3)/2
dV/dx = 125/2- 3x^2/2
125/2 = 3x^2/2 (to find station point)
125 = 3x^2
125/3 = x^2
x = sqrt(125)/sqrt(3)

second derivative = - 3x
when you put x = sqrt(125)/sqrt(3), second der = negative meaning that there is maximum volume at this point.

sub x in to find corresponding y value and find the volume then.

NEXT Q:

11. A soft drink manufacturer wants to minimise the amount of aluminium in its cans while still holding 375 mL of softdrink. Given that 375 mL has a volume of 375 cm3,
a) Show that the SA of a can is given by S = 2(pi)r2 + 750/r
b) Find the radius of the can that gives the min SA

This quesiton is basically same as the one above. Find the equation for volume and get the y and x values to find the equation for SA.

Makro · Mar 1, 2009

lyounamu said:
5. The height in metres of an object thrown up into the air is given by h = 20t - 2t2 where t is time in seconds. Find the max height that the object reaches.

dh/dt = 20 - 4t
t = 5 when dh/dt = 0 (i.e. to find a stationary point)

second derivative = -4
meaning that t=5 is a maximum point

sub t= 5

h = 20 x 5 - 2.5^2 = 50

Thanks, rather easy.

13. A rectangular prism with a square base is to have a surface area of 250 cm2.
a) Show that the volume is given by V = (125x - x3)/2
b) Find the dimensions that will give the max volume.

let height be y and base be x.

so 2x^2 + 4xy = 250
4xy = 250 - 2x^2
y = (250-2x^2)/4x

V = x^2 . y = (125x - x^3)/2
dV/dx = 125/2- 3x^2/2
125/2 = 3x^2/2 (to find station point)
125 = 3x^2
125/3 = x^2
x = sqrt(125)/sqrt(3)

second derivative = - 3x
when you put x = sqrt(125)/sqrt(3), second der = negative meaning that there is maximum volume at this point.

sub x in to find corresponding y value and find the volume then.

Are you saying sub x = sqrt(125)/sqrt(3) into the SA equation or V equation? I'm not sure which one. Even then how are you supposed to do it with all the roots? Answers say the dimensions are 6.45 x 6.45 x 6.45.

NEXT Q:

11. A soft drink manufacturer wants to minimise the amount of aluminium in its cans while still holding 375 mL of softdrink. Given that 375 mL has a volume of 375 cm3,
a) Show that the SA of a can is given by S = 2(pi)r2 + 750/r
b) Find the radius of the can that gives the min SA

This quesiton is basically same as the one above. Find the equation for volume and get the y and x values to find the equation for SA.

I'll wait for your answer to my question before attempting this

Thanks for the effort.

lyounamu · Mar 1, 2009

Makro said:
Are you saying sub x = sqrt(125)/sqrt(3) into the SA equation or V equation? I'm not sure which one. Even then how are you supposed to do it with all the roots? Answers say the dimensions are 6.45 x 6.45 x 6.45.

Well my answer is correct.

x = sqrt (125)/sqrt(3) = 6.4549...which is one of the dimension.

If you sub x value into y, you will find the y value as well.

Since x is the base of the square base, the dimension is somethin like:

6.45 x 6.45 x y (and you can find the y value by subbing x value in it)

I just subbed y value, and found that y value is actually 6.454972244 which is same as x.

Makro · Mar 1, 2009

Oh rightt, sorry and thanks

I got it. Now to attempt the 3rd one all by myself.

Makro · Mar 1, 2009

There are no x and y values for Q11, I got a) done and just don't know what do with b). Help me, please.

Also:

a 5 mn length of timber is used to border a triangular garden bed, with a the other sides of the garden against theh ouse walls.
a) show that the area of the garden is A = 1/2x P(sqrt)(25-x²)
b) Find the greatest possible area of the garden bed.

Now what are we supposed to do, it only gives me the length of one said that is neither x nor y. What do I do?

lyounamu · Mar 1, 2009

11. A soft drink manufacturer wants to minimise the amount of aluminium in its cans while still holding 375 mL of softdrink. Given that 375 mL has a volume of 375 cm3,
a) Show that the SA of a can is given by S = 2(pi)r2 + 750/r
b) Find the radius of the can that gives the min SA

v = pi r^2 h = 375
so h = 375/(r^2 pi)

A = 2pi r^2 + 2pi r h = 2 pi r^2 + 750/r ( if you sub h in)

dA/dr = 4pi r - 750/r^2
so 4pi r = 750/r^2

i.e. pi r ^3 = 375/2
r^3 = 375/2pi

r = 3.90796....

second deri = 4pi + 1500/r^3 so if you sub r value, you get positive 2nd derivative, meaning that it is the minimum...

so r = 3.90796...

lyounamu said:
if you don't have x and y value, you create them or at least you treat them as x or y. so i treat A as my y and r as my x here.

lyounamu · Mar 1, 2009

Makro said:
There are no x and y values for Q11, I got a) done and just don't know what do with b). Help me, please.

Also:

Now what are we supposed to do, it only gives me the length of one said that is neither x nor y. What do I do?

Can you write your question again?

I find it hard to read it.

Makro · Mar 1, 2009

Can you write your question again?

I find it hard to read it.

You answered it. You said pretend they're x and y values, as in make x = r and A = h. Also is A = Area?

Thank you. I'm interested in the working out of this in particular:

dA/dr = 4pi r - 750/r^2

And then

so 4pi r = 750/r^2

i.e. pi r ^3 = 375/2

Where does pi r³ come from?

Sorry for the annoying questions.

lyounamu · Mar 1, 2009

Makro said:
Thank you. I'm interested in the working out of this in particular:

well, i made that equal to zero so:

4pi r = 750/r^2
r^3 = 375/2pi

Makro · Mar 1, 2009

Thank you for that question.

Now I've confused myself with the last one I posted, let's just start fresh. [hr]

a 5 m length of timber is used to border a triangular garden bed, with the other sides of the garden against the house walls.
a) show that the area of the garden is A = 1/2x P(sqrt)(25-x2)
b) Find the greatest possible area of the garden bed.

I've done a). How are we supposed to do b)?

I've differentiated the eq. given in part a) and end up with -5x, but I don't know what it's supposed to equal? I must be missing something :/

lyounamu · Mar 1, 2009

Makro said:
Thank you for that question.

Now I've confused myself with the last one I posted, let's just start fresh. [hr]

I've done a). How are we supposed to do b)?

I've differentiated the eq. given in part a) and end up with -5x, but I don't know what it's supposed to equal? I must be missing something :/

But what exactly is P(sqrt)???

Makro · Mar 1, 2009

lyounamu said:
But what exactly is P(sqrt)???

I must have made a typo, sorry. Thought I checked it.

A = 1/2x(sqrt)(25-x2)

renny 123 · Mar 1, 2009

lyounamu said:
5.

13. A rectangular prism with a square base is to have a surface area of 250 cm2.
a) Show that the volume is given by V = (125x - x3)/2
b) Find the dimensions that will give the max volume.

let height be y and base be x.

so 2x^2 + 4xy = 250
4xy = 250 - 2x^2
y = (250-2x^2)/4x

V = x^2 . y = (125x - x^3)/2
dV/dx = 125/2- 3x^2/2
125/2 = 3x^2/2 (to find station point)
125 = 3x^2
125/3 = x^2
x = sqrt(125)/sqrt(3)

second derivative = - 3x
when you put x = sqrt(125)/sqrt(3), second der = negative meaning that there is maximum volume at this point.

.

I have a test on this tomorrow! Im was just doing this question and im just not sure how you got y=(125x - x^3)/2 from y = (250-2x^2)/4x
Sorry this might be a stupid question..

lyounamu · Mar 1, 2009

renny 123 said:
I have a test on this tomorrow! Im was just doing this question and im just not sure how you got y=(125x - x^3)/2 from y = (250-2x^2)/4x
Sorry this might be a stupid question..

x^2 . y = (125x - x^3)/2

. means times

renny 123 · Mar 1, 2009

lyounamu said:
x^2 . y = (125x - x^3)/2

. means times

Its alright i figured it out, i was doing something stupid haa
thanks

lyounamu · Mar 1, 2009

Makro said:
I must have made a typo, sorry. Thought I checked it.

A = 1/2x(sqrt)(25-x2)

can you clearify again?

is it 1/(2x) or 1/2 . x???

Makro · Mar 3, 2009

lyounamu said:
can you clearify again?

is it 1/(2x) or 1/2 . x???

My bad once again, "1/2 . x"

Makro · Mar 5, 2009

Sorry, can anyone answer the question? I'll do it here so it's in one place.

a) Show that the are of the garden is $A = \frac{1}{2}x\sqrt{25 -x ^{2}}$
b) Find the greatest possible area of the garden bed.

I've done part a), I didn't know what to do with b). Now everyone should be able to read the equation because I'm not failing.

kurt.physics · Mar 7, 2009

Makro said:
Sorry, can anyone answer the question? I'll do it here so it's in one place.

a) Show that the are of the garden is $A = \frac{1}{2}x\sqrt{25 -x ^{2}}$
b) Find the greatest possible area of the garden bed.

I've done part a), I didn't know what to do with b). Now everyone should be able to read the equation because I'm not failing.

$A = \frac{1}{2}x\sqrt{25 -x ^{2}}$

$$let$\,\, u = \frac{1}{2}x \,\,$and$\,\, v= \sqrt{25 -x ^{2}}$

$$Then$,$

$\frac{du}{dx} = \frac{1}{2}$

$with$\,\, $the$\,\,$derivative$\,\,$of$\,\,v\,\,$we$\,\,$must$\,\,$use$\,\,$the$\,\,$chain$\,\,$rule$

$$let$\,\, n=25-x^2,\,\,$then$\,\,v=n^{\frac{1}{2} }$

$\frac{dv}{dx} = \frac{dv}{dn} \times \frac{dn}{dx}$

$\frac{dv}{dn} = \frac{1}{ 2\sqrt{25-x^2} }$

$\frac{dn}{dx} = -2x$

$\therefore \frac{dv}{dx} = -\frac{x}{\sqrt{25-x^2} }$

$$Now$,\,\,$

$$if$\,\, y=uv,\,\, $then$,$

$\frac{dy}{dx} = u\frac{dv}{dx} + v\frac{du}{dx}$

$\frac{dy}{dx} = \frac{1}{2}x \times -\frac{x}{\sqrt{25-x^2} } + \sqrt{25 -x ^{2}} \times \frac{1}{2}$

$\frac{dy}{dx} = -\frac{x^2}{2\sqrt{25-x^2} } + \frac{ \sqrt{25 -x ^{2}}}{2}$

$\frac{dy}{dx} = -\frac{x^2}{2\sqrt{25-x^2} } + \frac{25 -x^2}{2\sqrt{25-x^2} }$

$\frac{dy}{dx} = \frac{25 -2x^2}{2\sqrt{25-x^2} }$

$$For stationary points$\,\, \frac{dy}{dx} = 0$

$\frac{25 -2x^2}{2\sqrt{25-x^2} } = 0$

$25 -2x^2 = 0$

$2x^2 = 25$

$x^2 = \frac{25}{2}$

$x = \pm\sqrt{ \frac{25}{2} }$

$x = \pm\frac{5}{\sqrt{2} }$

$x = \pm\frac{5\sqrt{2} }{2}$

$As x is the side of the garden, then we may only take the positive value. I will assume this is a maximum value as that is what this whole question is about.$

$\therefore A_{max} = \frac{1}{2}x\sqrt{25 -x ^{2}}$

$A_{max} = \frac{25}{4}$

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