# Thread: Optimization. (aplications of differentiation)

1. ## Optimization. (aplications of differentiation)

Alright..
1) Find the dimension of the rectangle of the largest area that can be inscribed in an equilateral triangle of side L if one side of the rectangle lies on the base of the triangle.

Ok... so, for the rectangle... the height will be y... and the length is x..
Area = XY, Perimeter = 2x+2y, etc.

For the triangle, the base is B, and ive been told the height is (Radical3 over 2)times B. Why is this true? I think i forgot my triangles...

and after that, where do i go then to get to the max area.

2) A cone-shaped drinking cup is made from a circular piece of a paper of radius R by cutting out a sector and joining the angles CA and CB. Find the maximum capacity of such a cup.

and lastly, 3) A cone with height h is inscribed in a larget cone with a height H so that its vertex is at the center of the base of the larger cone. show that hte inner cone has a maximum volume when h = (1/3)H

If you can show me how to work these specific problems Itd be greatly appreciated. I have a test tomorrow and the more information I know, the better.

Thanks!

2. A cone in a cone. Different than the usual cone in a sphere, cylinder in a cone, etc. Have not seen this one before. But, I would say it is tackled like the other ones......Similar triangles.

$\frac{r}{H-h}=\frac{R}{H}$

Solve for $r=\frac{R(H-h)}{H}$

Sub into the formula for cone volume:

$V=\frac{1}{3}{\pi}(\frac{R(H-h)}{H})^{2}h$

Differentiate:

$\frac{dV}{dh}=\frac{(h-H)(3h-H)R^{2}{\pi}}{3H^{2}}$

Now, set to 0 and solve for h:

We get $h=\frac{H}{3}$

3. For the rectangle in the equilateral triangle we can use similar triangles again.

The rectangle has dimensions 2x and y. The area is 2xy.

$\frac{\frac{\sqrt{3}}{2}L}{\frac{L}{2}}=\frac{y}{\ frac{L}{2}-x}$

Which reduces to: $\sqrt{3}=\frac{-2y}{2x-L}$

Solve this for $y=\frac{-\sqrt{3}(2x-L)}{2}$

Sub into the area formula: $2x(\frac{-\sqrt{3}(2x-L)}{2})=-\sqrt{3}x(2x-L)$

Now, differentiate wrt x, set to 0 and solve for x. y will follow and the area is in the bag.

4. Let h and r be the dimensions as in the diagram.

The volume is then $\frac{1}{3}{\pi}r^{2}h$.

But, $r^{2}+h^{2}=L^{2}$

therefore, whence and hence, $r^{2}=L^{2}-h^{2}$

$V=\frac{1}{3}{\pi}(L^{2}-h^{2})h$

$=\frac{1}{3}{\pi}(L^{2}h-h^{3})$,

for $0\leq{h}\leq{L}$.

$\frac{dV}{dh}=\frac{1}{3}{\pi}(L^{2}-3h^{2})$

Set to 0 and solve for h, we get $h=\frac{L}{\sqrt{3}}$

If $h=\frac{L}{\sqrt{3}}$, then $V=\frac{2\pi}{9\sqrt{3}}L^{3}$.

So the volume is as large as possible when $h=\frac{L}{\sqrt{3}}, \;\ and \;\ r=\sqrt{\frac{2}{3}}L$

Now, if r and h is the radius of the cone, the slant height of the cone will be

R, the radius of the circular sheet we are cutting the sector from.

Therefore, the largest volume is $\boxed{\frac{2\pi}{9\sqrt{3}}R^{3}}$