A little Caluclus help pls?

[LuDaCriS66]

I've got a couple of questions I need a little help with. Not too sure about these calculus problems. Any help is appreciated.

The chapter is on Extreme Values.

1) A printed page is to contain 60 cm^2 of printed material with clear margins of 5 cm on each side and 3 cm on the top and bottom. Find the minimum total area of the page.

Diagram

2) The illumination of an object by a light source is directly proportional to the strength of the source and inversely proportional to the square of the distance from the source. Two streetlights are 40 m apart and one is twice as strong as the other. Where is the darkest spot between the two lights?

For #2, if it requires a diagram... just forget it then... I'll see what I can do about it

3) Two towns A and B are 5km and 7km, respectively, from a railroad line. The points C and D nearest to A and B on the line are 6 km apart. Where should a station be located to minimize the length of a new road from A to S to B?

Diagram question #2

Thanks for any help.

 

[LuDaCriS66]

bump

 

[bigalt]

<< The chapter is on Extreme Values. >>


Sounds like they're trying to make calculus hip for the new generation.

 

[Oscar1613]

#1) let x= inner length and y=inner height and A=total area

xy=60
so y=60/x

A=(x+10)(y+6)
substitute for y
A=(x+10)(60/x + 6)
A=60 + 6x + 600/x + 60
A=6x + 600/x + 120

dA/dx=6 - 600/x^2
factor out an 6x^-2:
dA/dx=6x^-2(x^2-100)


if you cant figure it out from there... good luck in the rest of calculus

 

[Mean MrMustard]

3.8571426 miles from point C...but I could be wrong

 

[eLiu]

Hey

Optimization problems...YUCK.

I can help you set up #3 and #1, but I'm not too sure on #2. Evaluating will be up to you.

#1: you want to minimize area...so we can set up the basic eqn form: A=xy, where x=the inside length and y=inside width (inside as in not including the margins)

Now...since we want the area with the margins...A=(x+10)(y+6). Since xy=60, we can say that y=60x. Substitute, and evaluate.

we have that A=stuff. Now, in order to find the minimum, we need to differentiate and find/identify critical values. (by now, you should know howto ID max/mins).

For #3: I can set this up, but I'm not gonna bother to evaluate.

Anyways, in order to minimize the lengths of the roads, we'll want to minimize angle ASB (theta).
2 ways to do this...minimize theta, or maximize the 2 angles beside it...trust me, the 2nd method is easier by far. Of course, we'll want to maximize the angles in terms of X (trigonometry fun!)

So: let's call angle ASC alpha, and angle BSD beta, which will equal 180-theta-alpha. Since we want to maximize both at the same time, let's call their sum Z.
To get the angles in terms of X (define CS as x, and SD as 6-x):
tan(alpha)=5/x, so alpha=arctan(5/x)
tan(beta)=(7/6-x), so beta=arctan(7/6-x)

Now, Z=alpha=arctan(5/x) + beta=arctan(7/6-x). Differentiate & find the max or min.

Lemme think about #2...if you have questions, IM me (SunTzu86) or whatever.

 

[eLiu]

oo goodie...ur in luck, I managed to figure out howto setup #2 as well

First, translate that garbage into math symbols: I=illumination, S=strength, and d=distance
I=ks/d^2, where K is an arbitrary constant.

My diagram is drawn like this...teh 2x strength light is on te left, and the regular strength is on the right. D1 represents the distance to the dark point from 2x light, 40-D1 is the dist from the other light.

In order to find the darkest point, we must minimize the Ilumination there.
Basic eqn: Total illumination=I1 + I2
I1=k2s/d1^2
I2=ks/(10-d1)^2
So, Total=k2s/d1^2 + ks/(10-d1)^2
Note: k and s are both constant values...when u differentiate, don't forget that

Anyways, now that you have total in terms of a single var, minimize it--differentiate, set equal to 0, find critical values, blah.

Make sure you watch your chain rule factors, and note that the k and s will get divided out (so don't worry about them).

have fun,
-erc

 

[LuDaCriS66]

Thanks a lot! I really appreciate it! Thanks especially for your explanations.

 

[eLiu]

<< Thanks a lot! I really appreciate it! Thanks especially for your explanations. >>



My pleasure Good review for me...evil evil evil evil AP tests coming up...*shudder*

-Eric

 

[Oscar1613]

<< My pleasure Good review for me...evil evil evil evil AP tests coming up...*shudder*

-Eric >>



eh... they cant be that bad... i mean getting the right answer is only 1 point... the rest is set up and process... im not too worried about it... not to mention my teacher makes her regular tests harder than the AP exam... :Q