PLZ HELP ME!!?

Question

AP Math Calculus Problems?
PROBLEM 1 : Find two nonnegative numbers whose sum is 9 and so that the product of one number and the square of the other number is a maximum.
Click HERE to see a detailed solution to problem 1. 




PROBLEM 2 : Build a rectangular pen with three parallel partitions using 500 feet of fencing. What dimensions will maximize the total area of the pen ?
Click HERE to see a detailed solution to problem 2. 




PROBLEM 3 : An open rectangular box with square base is to be made from 48 ft.2 of material. What dimensions will result in a box with the largest possible volume ?
Click HERE to see a detailed solution to problem 3. 




PROBLEM 4 : A container in the shape of a right circular cylinder with no top has surface area 3 ft.2 What height h and base radius r will maximize the volume of the cylinder ?
Click HERE to see a detailed solution to problem 4. 




PROBLEM 5 : A sheet of cardboard 3 ft. by 4 ft. will be made into a box by cutting equal-sized squares from each corner and folding up the four edges. What will be the dimensions of the box with largest volume ?
Click HERE to see a detailed solution to problem 5. 




PROBLEM 6 : Consider all triangles formed by lines passing through the point (8/9, 3) and both the x- and y-axes. Find the dimensions of the triangle with the shortest hypotenuse.
Click HERE to see a detailed solution to problem 6. 




PROBLEM 7 : Find the point (x, y) on the graph of   nearest the point (4, 0).
Click HERE to see a detailed solution to problem 7. 




PROBLEM 8 : A cylindrical can is to hold 20 m.3 The material for the top and bottom costs $10/m.2 and material for the side costs $8/m.2 Find the radius r and height h of the most economical can.
Click HERE to see a detailed solution to problem 8. 




PROBLEM 9 : You are standing at the edge of a slow-moving river which is one mile wide and wish to return to your campground on the opposite side of the river. You can swim at 2 mph and walk at 3 mph. You must first swim across the river to any point on the opposite bank. From there walk to the campground, which is one mile from the point directly across the river from where you start your swim. What route will take the least amount of time ?
Click HERE to see a detailed solution to problem 9. 




PROBLEM 10 : Construct a window in the shape of a semi-circle over a rectangle. If the distance around the outside of the window is 12 feet, what dimensions will result in the rectangle having largest possible area ?
Click HERE to see a detailed solution to problem 10. 




PROBLEM 11 : There are 50 apple trees in an orchard. Each tree produces 800 apples. For each additional tree planted in the orchard, the output per tree drops by 10 apples. How many trees should be added to the existing orchard in order to maximize the total output of trees ?
Click HERE to see a detailed solution to problem 11.

Answer

What's with the Click here to see solution thing

PROBLEM 1 :
okay so let the numbers be x and y, where 9 &ge; x &ge; y &ge; 0
x + y = 9
or 
y = 9 - x
and 
Z = yx^2

Subtitute y to get
Z = 9x^2 - x^3
To find maximum, we must find dZ/dx and d^2 Z/dx^2
dZ/dx = 18x - 3x^2
= 3x(6 - x)

d^2 Z/dx^2 = 18 - 6x
= 6(3 - x)

From this, we conclude there's a stationary point at x = 0 and x = 6, and the concavity is negative when x > 3 therefore the maximum value for yx^2 is when x = 6
Therefore the numbers are
x =6 and y = 3

PROBLEM 2 :
3 divisions means two internal fence, assuming the divisions are also rectagular
so, let the dimensions be x and y, where the internal fences are aligned in the same direction as x
5x + 2y = 500
or
x = 100 - 2y/5

and
A = xy

substitute x to get
A = 100y - 2y^2 /5

Now we find the maxima
dA/dy = 100 - 4y/5
= 4/5(80 - y)

so yeah, substitude y = 80 to get x
and there you have it
the dimensions are
68 feet by 80 feet

in case of non rectangular divisions... i really don't know


PROBLEM 3 :
Okay, so let's assign some dimensions to the box
Since the base is a square, one of the dimension is the same as the other
Let them be x, x and h
anyway, the surface area of the open box is
x^2 + 4 xh = 48
solving for h gives
h = 12/x - x/4
The volume is
V = hx^2 

Substitute h to get
V = 12x - x^3 /4

Anyway, the derivative is
dV/dx = 12 - 3x^2 /4 
= 3/4 (4 - x)(4 + x)

Obviously dimension has to be positive so substitute x = 4 to get h
The dimensions are 4  feet by 4 feet by 2 feet

PROBLEM 4 :
Okay so
V = pi r^2 h
and
pi r^2 + 2 pi r h = 3

anyway, same procedure
h = 3/2pir - r/2
V = 3r/2 - pi r^3 /2
dV/dr = 3pi/2 (1/&radic;pi - r)(1/&radic;pi + r)

r = 1/&radic;pi
h = 1/&radic;pi

PROBLEM 5 :
Okay, so the squares of size x will subtract from the base by 2x (because there's two squares on each side)
So let the dimensions be
h = x, b = 3 - 2x, l = 4 - 2x
V = (3 - 2x)(4 - 2x)x
= 4x^3 -14x^2 +12x

dV/dx = 12x^2 -28x +12
= 12(x^2 - 7x/3 +1)

d^2 V/dx^2 = 24x - 28

wow, this is ugly
x = (7 + &radic;13)/6

the dimensions are
2.2 foot by 1.2 foot by 1.8 foot
(1 decimal place)


Wow, that was a lot, I'll answer the rest later
Hope you found this helpful
(i really deserve more recognition from this)
(you should've split the question into multiple questions)

Answer

Technically since question 6 didn't specify that the triangle has to be in the first quadrant, one could make the line go though the origin and end up with a hypotenuse of size 0

I'll try to find a solution for if it was restricted though, just, this is not easy

Also, problem 7 is not a valid question

Answer

PROBLEM 8 : 
20 = pi r^2 h
h = 20/(pi r^2) 

SA = 2 pi r^2 + pi r h
$ = 20pi r^2 + 8pi r h
= 20pi r^2 + 160/r

d$/dr = 40pi r - 160/ r^2
Set this to 0 and solve 

0 = 40 pi r^3 - 160
4/pi = r^3
r = (4/pi)^(1/3) = 1.1 ,  r must be a real number because it represents a real quantity
h = 10/cuberoot(2pi) = 5.4


PROBLEM 9 : requires more information on which direction the camp lies from with respect to the river, I'll try and make it up later


PROBLEM 10 : 
let's say the rectangle has a length of 2r and a height of h
20 = 2r + 2h + pi r
h = 10 - (2 +pi)r/2

A = 2rh
= 20r - (2 +pi)r^2

da/dr = 20 - 2r(2 + pi)
= 2(2 + pi)(10/(2 + pi) - r)
the dimension (of the rectangle) is 20/pi+2 = 3.9 by 5


PROBLEM 11 :
So yeah, the total output is
O = 10(50+n)(80 - n)
= 40000+300n-10n^2

dO/dn = 300 - 20n
= 20(15 - n)

so, plant 15 more trees

PLZ HELP ME!!?

Responses (4)