4 Maxima and Minima Qu's (1 Viewer)

[currysauce]

Hi all, i need help with 4 questions that i am trying to work through, out of this 3 unit book (which has harder 2u) for maxima and minima...that i found in my library

if anyone could answer any of these questions it would be greatly appreciated... i've just lost myself trying to find formulas

anywho thanks for replying

1. A surfboard is in the shape of a rectangle and a semicircle. The perimeter is to be 4m. Find the max. area of the surfboard, to 2.d.p.

2. Find the least surface area, to the nearest cm^2, of a closed cylinder that will hold a volume of 400cm^3.

3. A 3m piece pf wore os cut into 2 peices and bent around to form a square and a circle. Find the size of the two lengths, correct to 2 deimal places, that will make the total area of the square and circle a minimum.

4. The FRAME of a picture has a border of 2cm at the top and bottom and 3cm at the sides. If the total area of the border is to be 100cm^2, find the maximum area of the frame.

Now i'll put the answers from the book down.

1. 1.12
2. 301
3. 1.68, 1.32
4. 160and a 6th

THANKS SO MUCH

 

[Trev]

answer to q2:
surface area cyl. = 2πr² + 2πrh (r - radius, h - height)
volume cyl. = πr²h

πr²h = 400
h = 400/πr²

therefore;
surface area cyl. = 2πr² + 2πr[400/πr²]
SA = 2πr² + 800/r
(SA)' = 4πr - 800/r²
let (SA)' = 0 for stationary pts.
therefore;
4πr - 800/r² = 0
πr³ = 200
r = 3(sqrt)[200/r] (the third root of 200/r)
cant b bothered showing that its a min pt, u can just tell it is.
then sub r into 2πr² + 800/r and uve got the min surface area

 

[alien]

answer to q2:
surface area cyl. = 2πr² + 2πrh (r - radius, h - height)
volume cyl. = πr²h

πr²h = 400
h = 400/πr²

therefore;
surface area cyl. = 2πr² + 2πr[400/πr²]
SA = 2πr² + 800/r
(SA)' = 4πr - 800/r²
let (SA)' = 0 for stationary pts.
therefore;
4πr - 800/r² = 0
πr³ = 200
r = 3(sqrt)[200/r] (the third root of 200/r)
cant b bothered showing that its a min pt, u can just tell it is.
then sub r into 2πr² + 800/r and uve got the min surface area

i can't believe you just did that! school is supposed to be over! that means no more thinking! (unless you didn't just finish year12)

currysauce, why don't you ask your teacher? it's usually easier to learn something if someone tells you face to face. unless it's urgent but really

 

[Trev]

answer question 4

area of border is 2.2.x + 3.2.y - (4.2.3)
area of broder = 4x + 6y - 24 = 100
= 2x + 3y = 62
therefore y = (62 - 2x)/3

since area frame is xy then..
A = xy
A = x[(62 - 2x)/3]
= 62x/3 - 2x²/3
A' = 62/3 - 4x/3
for stationary pts let A' = 0
0 = 62/3 - 4x/3
4x = 62
x = 15.5, therefore from y = (62 - 2x)/3; y = 10 1/3

therfore max area of frame is xy = 15.5 by 10 1/3
= 160 1/3cm^2

 

[Trev]

i can't believe you just did that! school is supposed to be over! that means no more thinking! (unless you didn't just finish year12)

currysauce, why don't you ask your teacher? it's usually easier to learn something if someone tells you face to face. unless it's urgent but really

im doin yr 12 now, i cant b bothered studying so i thought if i answer this q itll b study lol, coz my first ext1 HSC assessment is on tues, and i havnt started studying so yeh...... lol

 

[alien]

im doin yr 12 now, i cant b bothered studying so i thought if i answer this q itll b study lol, coz my first ext1 HSC assessment is on tues, and i havnt started studying so yeh...... lol

o ok. good luck then!

 

[Trev]

answer quesiton 3:

square will use 4x, therefore circle will use 3 - 4x (soz not good at explaining, this is just wat ive got in my book)
4x perim square, 3-4x circum circle = 2.pi.r
2.pi.r = 3-4x
r = (3-4x)/2.pi

area square = x^2, area circle = pi.r^2 (soz cant b bothered using the symbols lol)
therfore A = x^2 + pi.r^2
A = x^2 + π[(3-4x)/2π]²
A = x² + (9 - 24x + 16x²)/4π
therfore
A' = -6/π + 8x/π + 2x
A' = 0 for stationary pts
0 = -6 + 8x + 2xπ
x = 3/(4 + π)

therfore lengths are 4x = 1.68, 3-4x = 1.32

 

[withoutaface]

Q1:

ok let the length of the rectangle be x, and breadth be y

Now the total perimeter is equal to 2x+y+pi*y/2=4 (circumference of circle=pi*diameter)

Now area is given by A=xy+pi*y²/8

Which, when we find x in terms of y goes to:

y(2-y/2-pi*y/4)+pi*y²/8=A
2y-y²/2-pi*y²/8=A
dA/dy=2-y-pi*y/4
d²A/dy²=-1-pi/2

Equate this to 0 and find the value of y for which A is a maximum

y+pi*y/4=2
y=2/(1+pi/4)

And substitute this value of y into the area equation and this should give you the maximum area

 

[currysauce]

thanks so much guys for giving up ur time

 

[Numero Uno]

heh is this the Maths in Focus Book?

 

[Trev]

yeh its in the math in focus hsc textbook

 

[~ ReNcH ~]

yeh its in the math in focus hsc textbook

Hated that book!

 

[Numero Uno]

Hated that book!

Yes ... its gay