Max and Min Problems (Differentiation) (1 Viewer)

[atar90plus]

Hello
I am wondering if you guys can help me with these max and min problems. Please show full working out and answers.

6. A closed cylindrical can is to have a capacity of 16 pi cm^3 . What are the radius of the base and the height of the cylinder for the total surface area to be a minimum?

11. A printed page of a book is to have a side margin 1cm, a top margin of 2cm and a bottom margin of 3cm. It is to contain 200cm^2 of printed matter. Find the dimensions of the page if the area of paper used is to be a minimum.

22. An enclosure of area 200m^2 is to be made in the shape of a rectangle with an existing wall as one side of the rectangle.

a) If one side of the rectangle is x, show that the length L metres required is given by = 2x + 200/x

b) Find the minimum length of fencing required

 

[Arr0gance]

 

[bobmcbob365]

11. A printed page of a book is to have a side margin 1cm, a top margin of 2cm and a bottom margin of 3cm. It is to contain 200cm^2 of printed matter. Find the dimensions of the page if the area of paper used is to be a minimum.



Let x = width of the printed matter.
Let y = height of the printed matter.

So, Area of printed matter (area of rectangle) : xy = 200
Let this be equation 1.

The overall width of the printed page is x + the two margins.
x + 1 + 1
= x + 2
The overall height of the printed page is x + the two margins.
y + 2 + 3
= y + 5

So, Area of overall page (area of rectangle) : A = (x+2)(y+5)
Let this be equation 2.

Solving equation 1 for y:
y = 200/x

Sub this into equation 2:
A = (x+2)((200/x)+5)

Expand and simplify:
A = 5x + (400/x) + 210

Differentiate in respect to x:
A' = 5 - (400/x^2)
Let A' = 0:
5 - (400/x^2) = 0
5x^2 - 400 = 0
x^2 = 80
x = 4*sqrt(5)

Substitute this back into equation 1:
Gives:
y = 200 / (4*sqrt(5))
y = 10 sqrt(5)

Since width of overall page is x + 2:
Sub in x:

Therefore, Width = 4sqrt(5)+2

Since height of overall page is y + 5:
Sub in y:

Therefore, Height = 10sqrt(5) + 5

 

[_pikachu]

22) a) let the base of the rectangle be x
Area = Base x Height
200 = x(height)
height = 200/x

Perimeter = 2(height) + 2(base)
since there is already an existing wall, get rid of one of the sides
L = 2(base) + height
L = 2x + 200/x

b) L' = 2 -200/x^2
= 2(1-100/x^2)
L" = 400/x^3

At L' = 0
0 = 2-200/x^2
200/x^2=2
200=2x^2
100=x^2
x=plus or minus 10
since x>0 (x is a measurement)
x=10

sub into L''
L" = 400/10^3
L" = 0.4
L' > 0
therefore, minimum length

L = 2(10) + 200/10
L = 20 + 20
L = 40m
minimum length required is 40m

 

[_pikachu]

Also, i think this is 2unit, not extension 1