|
|
|
|
|
|
|
|
|
 Posted: Mon Nov 27, 2006 1:10 am
These are applications of the first and second derivative tests... Most are a breeze, but I am struggling with 3 of them... I would it if you would just help me get the correct formulas… (I am almost certain that is where I am messing up) Thank you for your help and may good luck go your way. number 1 A Norman window has the shape of a rectangle surmounted by a semicircle. (Thus, the diameter of the semicircle is equal to the width of the rectangle.) If the perimeter of the window is 30 ft, find the dimensions of the window so that the greatest possible amount of light is admitted. number 2 A piece of Wire 10 m long is cut into two pieces. One piece is bent into a square and the other is bent into an equilateral triangle. How should the wire be cut so that the total area enclosed is minimized? I actually have the answer to this one… That is how I know I am repeatedly doing it wrong… The answerCut wire at 40(3)^1/29+4(3)^1/2 for square number 3 A boat leaves a dock at 2:00 pm and travels due south at a speed of 20 km/h. Another boat has bee heading due east at 15 km/h and reaches the same dock at 3:00 pm. At what time were the two boats closest together?
|
 |
|
|
|
|
|
|
|
|
|
|
|
|
|
Posted: Mon Nov 27, 2006 5:35 am
The wire problem was my favorite so I'll answer it.
Let x be the length used for the square so 10-x is the length used for the triangle.
The area of the square is (x/4)^2=x^2/16. Note that it has to be divided by four since the square has four sides.
The area of the equilateral triangle is [[(10-x)/3]^2 sqrt(3)]/4=[(10-x)^2 sqrt (3)]/36 since the triangle has three sides and the formula for its area is [s^2 sqrt(3)]/4.
The areas of the square and triangle is x^2/16 + [[(10-x)/3]^2 sqrt(3)]/4.
Differentiating: dA/dx= 2x/16 + 2(10-x)(-1)(sqrt(3))/36= x/8 - (10-x)(sqrt(3))/18= 0 (Equate this to zero to get the maximum.)
This will result to x/8=(10-x)[sqrt(3)]/18;
x/8= 10[sqrt(3)]/18 - x[sqrt(3)]/18;
x/8 + x[sqrt(3)]/18= 10sqrt(3)/18;
x[1/8 + sqrt(3)/18]= 10sqrt(3)/18;
x= [10sqrt(3)][18] / [1/8 + sqrt (3)/18];
x= [10sqrt(3)][18] / [(9+4sqrt(3))/72];
x= 40sqrt(3) / [9+4sqrt(3)]
Now, I got the solution.
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
Posted: Mon Nov 27, 2006 6:14 am
Now, the window problem.
Let r be the radius of the semicircle and h be the height of the rectangle., and the width of the rectangle is 2r.
The restirction is the perimeter, which is equal to 3 sides (left, right and bottom) of the recatangle plus the perimeter of the semi-circle.
P=30= 2h + 2r + r(pi)
h=[30 - 2r -r(pi)]/2
The formula for the area is:
A=2rh + (pi)r^2/2
Substitute h=[30 - 2r -r(pi)]/2:
A=2r[30 - 2r -r(pi)]/2 + (pi)r^2/2= 30r - 2r^2 - (pi)r^2 + (pi)r^2/2
= 30r - 2r^2 - (pi)r^2/2
Differentiating: dA/dx= 30 - 4r - (pi)r = 0
r(pi) + 4r = 30
r= 30/(pi + 4) ft
2r=w= 60/(pi+4) ft
Just substitute the r in the equation: h= [30- 2r - r(pi)]/2
I am already sleepy and my mind can no longer work on the third problem right now.
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
Posted: Mon Nov 27, 2006 1:43 pm
3rd problem
Let t be time.
t=0 at 2.00PM
t=1 at 2.01PM
Let the dock be: x=0, y=0
x=1, y=0 is 1km east
x=0, y=-1 is 1km south.
At time t:
boat 1 is at x=t/3, y=0
boat 2 is at x=0, y=t/4-15
This means, their distance is exact: sqrt((t^2)/9+(t^2)/16-15*t/2+225) = sqrt(25*(t^2)/144-15*t/2+225) =: d
This reaches a minimum or maximum if it's derivate is 0. But, since it will always be positive, we can drop the sqrt.
derivate of d = 25*t/72-15/2
This is 0 when: t = 108/5 = 21.6 = 21 minutes and 36 seconds
So, answer is: at 2:21 and 36 seconds PM.
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
Posted: Mon Nov 27, 2006 9:36 pm
thank you guys for all of your help!
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
