Max Value problems (1 Viewer)

LoveHateSchool · Nov 14, 2011

A rectangular prism has a length which is twice its breadth. If the surface area is 48 cm^2, find the dimensions that will produce a maximum value.

And

A cylinder has a surface area of 160 cm^2, show that the volume is given by V=80r-πr³

Rep and cookies for anyone that helps the first one,not sure where I'm going wrong, and second I find it really hard to get out the formulas.

Alkanes · Nov 14, 2011

I'll do the second one.

V = $\pi r^2h$ (1)

S.A = $2\pi r^2 + 2\pi rh$ = 160

Make h the subject from the S.A and sub into V (1)

Alkanes · Nov 14, 2011

$\pi r^2 + \pi rh = 80$

H = $\frac{80 -\pi r^2}{\pi r}$

And yea should get the volume from here.

Sy123 · Nov 14, 2011

LoveHateSchool said:
A rectangular prism has a length which is twice its breadth. If the surface area is 48 cm^2, find the dimensions that will produce a maximum value.

And

A cylinder has a surface area of 160 cm^2, show that the volume is given by V=80r-πr³

Rep and cookies for anyone that helps the first one,not sure where I'm going wrong, and second I find it really hard to get out the formulas.

Do we get the height of the rectangular prism? (i.e. height=breadth)

LoveHateSchool · Nov 14, 2011

^Nope, only its surface area.

I assumed the breadth would have to be x and the length 2x and the height to be y and tried to work it out but it didn't work.

And when I tried subbing back into the V in the second one, I didn't get the formula they gave me to show

pokka · Nov 14, 2011

ok i'll answer the first one: let the height be h, the length be 2b and the breadth be b. Using these dimensions, you will find that the surface area is 6bh + 4b^2=48. Now, from here, make h the subject. [It should be in terms of b]
Now i'm assuming what you mean by "maximum value" you mean maximum volume. Well, you know that V=2b^2 h. Sub in the h you found into the equation then differentiate and from there you can find the maximum breadth and thus the maximum height as well (from the volume or surface area equation)
I hope this makes sense!

Alkanes · Nov 14, 2011

Lol are you srs? It's pretty much simple algebra work from there.

But anyways:

$V = \pi r^2h$

$\pi r^2(\frac{80-\pi r^2}{\pi r})$

$r(80-\pi r^2)$

$80r - \pi r^3$

SpiralFlex · Nov 14, 2011

ALWAYS draw a pretty diagram. Standard Spiral rule.

We need to find the surface area! We already know what it is so we need to form an equation to find the relationship between it and it's dimensions.

Easy!

$S=2*xy + 2*(2xy)+2*(2x^2)$

$4x^2+6xy=48$

$2x^2+3xy=24$

$x(2x+3y)=24$

$2x+3y=\frac{24}{x}$

$3y=\frac{24}{x}-2x$

$y=\frac{8}{x}-\frac{2}{3}x$

Now we need to construct the volume!

$V=x*2x*y$

$=2x^2(\frac{8}{x}-\frac{2}{3}x)$

$\therefore V=16x-\frac{4}{3}x^3$

$\frac{dV}{dx}=16-4x^2$

Do the second derivative for convenience,

$\frac{d^2V}{dx^2}=-8x$

For maximum or minimum, $\frac{dV}{dx}=0$

$16-4x^2=0$

$x=2$

We must test,

$\frac{d^2V}{dx^2}=-8(2)<0$

$\therefore$ When $x=2$ there will be a maximum.

$\therefore x=2$

$2x=4$

$y=\frac{8}{3}$

$V=80r-\pi r^3$

$\frac{dV}{dr}=80-3\pi r^2$

For maximum or minimum, $\frac{dV}{dr}=0$

$80-3\pi r^2=0$

$r=\pm \sqrt{\frac{80}{3\pi}}$ (But $r>0$ . We cannot have a negative distance!)

$\therefore r =\sqrt{\frac{80}{3\pi}}=2.91$

We must test.

$\frac{d^2V}{dr^2}=-6\pi r$

When $r=2.91$

$\frac{d^2V}{dr^2}=-6\pi (2.91)<0$

Therefore a maximum volume occurs when $r=2.91$

LoveHateSchool · Nov 14, 2011

Thank you so much Spi! It will never let me rep you enough!

I tried the further questions on the cylinder and I got the r to be 8.49, but that doesn't seem right because I get a minus value for the max volume!

SpiralFlex · Nov 14, 2011

LoveHateSchool said:
Thank you so much Spi! It will never let me rep you enough!

I tried the further questions on the cylinder and I got the r to be 8.49, but that doesn't seem right because I get a minus value for the max volume!

Edited with solutions!

LoveHateSchool · Nov 14, 2011

Thank you, I didn't take the square root, and hence there was the problem. I'm going to try the rest of the sheet now but I might need more help! Wish me luck!

Thank you Spi

LoveHateSchool · Nov 14, 2011

Ohh I did the rest of the Max and Min but now stuck on this.

A curve has second derivative=8x.
The tangent at the point (-2,5) is parallel to the line x-y+2=0. Find the equation of the curve.

I presume you have to find the primitive and then....any help please?

Alkanes · Nov 14, 2011

I get no thanks ?

LoveHateSchool · Nov 14, 2011

^Oh I repped you and said thanks on the rep comment! I couldn't rep Spi though, I should have thanked you in a tangible post too though, so I'm sorry.

Thank you Alkanes! You wonderful chemistry person!

SpiralFlex · Nov 14, 2011

LoveHateSchool said:
Ohh I did the rest of the Max and Min but now stuck on this.

A curve has second derivative=8x.
The tangent at the point (-2,5) is parallel to the line x-y+2=0. Find the equation of the curve.

I presume you have to find the primitive and then....any help please?

I fell asleep! My dad shut down my computer too! >_>

$y''=8x$

$y'=4x^2+C$

We know that at $x=-2$ . There is a tangent with a gradient parallel to that line, ie. 1.

$1=4(-2)^2+C$

$\therefore C = -15$

$\therefore y'=4x^2-15$

$y=\frac{4}{3}x^3-15x+D$

We know it passes through (-2, 5)

$5=\frac{4}{3}(-8)+30+D$

$\therefore D = -\frac{43}{3}$

The equation of the curve is,

$y=\frac{4}{3}x^3-15x-\frac{43}{3}$

LoveHateSchool · Nov 14, 2011

^I adore thee Spi and hope your own studies are faring well.

I know you can get 99+

But thank you for all the ongoing maths support, I will mention your name as one of the saviours of my maths scores when I get them

pokka · Nov 14, 2011

don't i get any thanks?

gwc · Jan 22, 2017

SpiralFlex said:
ALWAYS draw a pretty diagram. Standard Spiral rule.

We need to find the surface area! We already know what it is so we need to form an equation to find the relationship between it and it's dimensions.

Easy!

$S=2*xy + 2*(2xy)+2*(2x^2)$

$4x^2+6xy=48$

$2x^2+3xy=24$

$x(2x+3y)=24$

$2x+3y=\frac{24}{x}$

$3y=\frac{24}{x}-2x$

$y=\frac{8}{x}-\frac{2}{3}x$

Now we need to construct the volume!

$V=x*2x*y$

$=2x^2(\frac{8}{x}-\frac{2}{3}x)$

$\therefore V=16x-\frac{4}{3}x^3$

$\frac{dV}{dx}=16-4x^2$

Do the second derivative for convenience,

$\frac{d^2V}{dx^2}=-8x$

For maximum or minimum, $\frac{dV}{dx}=0$

$16-4x^2=0$

$x=2$

We must test,

$\frac{d^2V}{dx^2}=-8(2)<0$

$\therefore$ When $x=2$ there will be a maximum.

$\therefore x=2$

$2x=4$

$y=\frac{8}{3}$

For people that may look at this in the future, the final part is not the final answer. After doing that you have to sub x and y back into the volume equation to get the final answer of 21/1/3 cm^2

Max Value problems (1 Viewer)

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