Optimization Problems

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October 31st 2007, 07:13 PM
Super Mallow

Optimization Problems

Well, back with some more. I'm just not understanding them; I don't get very far with them. Currently, I have 2 I do not understand.

1) A rectangle is to be inscribed under the arch of the curve y=4 cos (.5X) from X= -Pi to Pi. What are the dimensions of the rectangle with the largest area, and what is the actual largest area?

2) A window is in the form of a rectangle surmounted by a semi-circle. The rectangle is of clear glass, whereas the semicircle is of tinted glass that transmits only half as much light per unit area as the clear glass does. The total perimeter is fixed. Find the proportions of the window that will admit the most light. Neglect the thickness of the frame

Any help is appreciated; I'm just not understanding how to do these
October 31st 2007, 07:49 PM
kalagota

Quote:

Originally Posted by Super Mallow

Well, back with some more. I'm just not understanding them; I don't get very far with them. Currently, I have 2 I do not understand.

1) A rectangle is to be inscribed under the arch of the curve y=4 cos (.5X) from X= -Pi to Pi. What are the dimensions of the rectangle with the largest area, and what is the actual largest area?
...
Any help is appreciated; I'm just not understanding how to do these

1)

$A=lw$
$w=4\cos{\frac{x}{2}}$
and
$l=2x \,$ due to symmetry.
hence
$A=8 x\cos{\frac{x}{2}}$
and derivative of A
$A' = -4x\sin {\frac{x}{2}} + 8\cos{\frac{x}{2}}$
then get the critical point, i.e. A' = 0, and solve for x, then those x are candidates for the solution. then use second derivative test evaluated at the value of x you got, and if it is negative, i.e. A"(x)<0, then that x is the solution.. take note that x should satisfy that it is between $\, -\pi \, and \, \pi$ Ü
October 31st 2007, 08:28 PM
Soroban

Hello, Super Mallow!

Quote:

2) A window is in the form of a rectangle surmounted by a semicircle.
The rectangle is of clear glass, whereas the semicircle is of tinted glass
that transmits only half as much light per unit area as the clear glass.
The total perimeter is fixed.
Find the proportions of the window that will admit the most light.

Code:

* * * * * * * * * r r * - - - - + - - - - * | | | | h| |h | | | | * - - - - - - - - - * 2r

The radius of the semicircle is $r.$
The width of the rectangle is $2r$ , its height is $h.$

The perimeter of the semicircle is: . $\pi r$
The perimeter of the rectangle is: . $2r + 2h$

The total perimeter is a constant, $P.$
Hence, we have: . $\pi r + 2r + 2h \:=\:P\quad\Rightarrow\quad h \:=\:\frac{P - \pi r - 2r}{2}$ . [1]

The area of the rectangle is: . $2rh$
It admits a certain amount of light per squre unit.
Its light admission is: . $L_1\:=\:2rh$

The area of the semicircle is: . $\frac{1}{2}\pi r^2$
It admits half as much light per square unit.
Its light admission is: . $L_2 \:=\:\frac{1}{2}\left(\frac{1}{2}\pi r^2\right) \:=\:\frac{1}{4}\pi r^2$

The total light admitted is: . $L \:=\:\frac{1}{4}\pi r^2 + 2rh$ . [2]

Substitute [1] into [2]: . $L \;=\;\frac{1}{4}\pi r^2 + 2r\left(\frac{P - \pi r - 2r}{2}\right)$

And we have: . $L \;=\;-\left(\frac{3\pi}{4} + 2\right)r^2 + Pr$

. . and that is the function we must maximize . . .
November 4th 2007, 09:06 AM
Super Mallow

Solutions look good! I have to still try the Window question, which I'll do in a minute

I do have a question for Kalagota, or anyone else who can answer it. I found the critical point (1.72, Approx SqreRoot 3), but now what do I do?

EDIT: I tried the window question...I don't get how to maximize the function because of the P in there
November 4th 2007, 07:20 PM
kalagota

Quote:

Originally Posted by Super Mallow

Solutions look good! I have to still try the Window question, which I'll do in a minute

I do have a question for Kalagota, or anyone else who can answer it. I found the critical point (1.72, Approx SqreRoot 3), but now what do I do?

EDIT: I tried the window question...I don't get how to maximize the function because of the P in there

assuming you are right, note that your critical point is in $(-\pi,\pi)$ . now, solve for the second derivative $A''(x)$ and plug in your critical point there, if $A''(x)<0$ , then you are done.. (you can now solve for your l and w)