# optimzation problem # 10

• August 13th 2009, 09:44 AM
skeske1234
optimzation problem # 10
A). The rectangle of maximum area that could be isncribed inside the triangle always has dimensions equal to half the lengths of the sides adjacent to the rectangle. Prove that this is true for any right triangle.

B) Prove that any cylindrical can of volume k cubic units that is to be made using a minimum amount of material must have the height equal to the diameter.
• August 13th 2009, 10:49 AM
skeeter
Quote:

Originally Posted by skeske1234
A). The rectangle of maximum area that could be isncribed inside the triangle always has dimensions equal to half the lengths of the sides adjacent to the rectangle. Prove that this is true for any right triangle.

B) Prove that any cylindrical can of volume k cubic units that is to be made using a minimum amount of material must have the height equal to the diameter.

I've already completed two problems like these ... you need to start showing some attempt.
• August 13th 2009, 12:01 PM
skeske1234
OK>>
I've been working on part 1 so far.. trying to, so i have:

I drew a triangle right angled, labelled points..
Triangle ABC has a rectangle in it labelled by EFD

B has the right angled and F is across from it. E is in the middle of the AB line and D is in the middle of the BC line.

Now for the points:
A(y,0)
E(0.5y,0)
D(0.5x,0)
C(x,0)

so y=mx+b
0=m and 0=b when i plugged them in

so my eqtn becomes y=x

Max Area of Rectangle
A(x)=xy
=x(x)
=x^2

A'(x)=2x
A'(x)=0 when x=0
A(0)=0

And from here, my proving did not go so well. ^.^ what should I do next?
• August 13th 2009, 01:02 PM
skeeter
this is just like that problem you did with the 5 and 12 sided legs of a right triangle.

let point A be on the y-axis with coordinates $(0,h)$

point B is at the origin

point C is on the x-axis with coordinates $(b,0)$

note that $b$ and $h$ are both positive constants

the equation of the line between A and C is $y = -\frac{h}{b}x + h$

F is the point on the line AC

BE = y ... BD = x

area of the rectangle is $R = xy$

area of the rectangle , $R = x\left(-\frac{h}{b}x + h\right)$

find $\frac{dR}{dx}$ and determine the value of x that maximizes R

you should get $x = \frac{b}{2}$ ... the x-value of the midpoint of AC
• August 13th 2009, 01:48 PM
skeske1234
For part B)

What does the question mean by the height must equal the diameter for the volume of a cylindrical can using the min amount of material?

How can the height equal the diameter? I'm looking at one of my previous questions at the moment and I can see that the radius is 43 so the diamater is 86 and the height is 172.. 172 does not equal 86.
(Thinking)

so what do they mean by this exactly?
• August 13th 2009, 01:51 PM
skeeter
Quote:

Originally Posted by skeske1234
For part B)

What does the question mean by the height must equal the diameter for the volume of a cylindrical can using the min amount of material?

How can the height equal the diameter? I'm looking at one of my previous questions at the moment and I can see that the radius is 43 so the diamater is 86 and the height is 172.. 172 does not equal 86.
(Thinking)

so what do they mean by this exactly?

h = 2r ... if you look at the cylinder from the side, it looks like a square.
• August 13th 2009, 03:16 PM
skeske1234
Quote:

Originally Posted by skeeter
h = 2r ... if you look at the cylinder from the side, it looks like a square.

Ok, can you check if I am doing it right so far?

so I said that

r^2+(0.5h)^2=[r^2+(0.5h)^2]^2
r^2=[16r^4+8r^2h^2+h^4-4h^2]/16

Max Volume Cylinder

V(r)=pi(r^2)h
=[16pi(h)^4+8pi(h^3)(r^2)+pi(h^5)-4pi(h^3)]/16
V'(r)=1024pi(h)(r^3)+256pi(h^3)(r)+80h^4-192pi(h^2)
• August 13th 2009, 04:14 PM
skeeter
Quote:

Originally Posted by skeske1234
B) Prove that any cylindrical can of volume k cubic units that is to be made using a minimum amount of material must have the height equal to the diameter.

$k = \pi r^2 h$

$h = \frac{k}{\pi r^2}$

minimize surface area ...

$S = 2\pi r^2 + 2\pi r h$

$S = 2\pi r^2 + 2\pi r \cdot \frac{k}{\pi r^2}
$

$S = 2\pi r^2 + \frac{2k}{r}$

$\frac{dS}{dr} = 4\pi r - \frac{2k}{r^2}
$

set $\frac{dS}{dr} = 0$

$4\pi r = \frac{2k}{r^2}$

$r^2 = \frac{2k}{4\pi r} = \frac{k}{2\pi r}$

$h = \frac{k}{\pi r^2} = \frac{k}{\pi} \cdot \frac{1}{r^2} = \frac{k}{\pi} \cdot \frac{2\pi r}{k} = 2r$