Question

Find a formula for

1) the area of the figure
2) the perimeter of the figure

And then find the dimensions of x and y that maximize the area given that the perimeter is 200.

write solutions in increasing order of x and enter the EXACT answers
x=____ ; y= _______

and

x=_____; y=______

@ganeshie8 ? :)

**different figure this time :)

Answer 1

figure :)

Answer 2

start by writing out expressions for Area and Perimeter

Answer 3

Area = ?
Perimeter = ?

Answer 4

umm area= 4 semi circles + 1 rectangle?

haha not sure :/ this one confuses me :/

Answer 5

yes,

if u look at it carefully,
its just "one rectangle" and "two full circles".

Answer 6

think of the top and bottom semicircles as broken pieces of same circle..

Answer 7

similarly, left and right semicircles also form one full circle.

Answer 8

ohhh didn't see it!! haha :P

so 2πr^2 * L * W ?

Answer 9

try to setup the expression for Area and Perimeter

Answer 10

Nooo... the radius is not same for all the circles

Answer 11

ohh then i don't know :(

Answer 12

the radius is same for top/bottom semicircles
the radius is same for left/right semicircles

Answer 13

Area of entire figure = ( rectangle) + (top and bottom semi circles) + (left and right semi circles)

Answer 14

okay:)
so how would that look? :/

Answer 15

Area of entire figure = \(\large (xy) + (\pi [\frac{x}{2}]^2) + (\pi [\frac{y}{2}]^2) \)

Answer 16

simplifying gives u :

Area of entire figure = \(\large xy + \frac{\pi x^2}{4} + \frac{\pi y^2}{4}\)

Answer 17

see if that looks okay

Answer 18

yes i think so :)

because the two circles are different right?

so what about perimeter? :/

Answer 19

see if u can find out

Answer 20

il give a hint : two circles

Answer 21

so would it be like this?

2πx + 2πy ?

Answer 22

almost right,

whats the radius of each circle ?

Answer 23

Is it x or x/2 ?

Answer 24

oh so

2πx 2πy
----- + -------
2 2
?

Answer 25

so hard

Answer 26

yes !

that gives :

Perimeter = \(\large \pi x + \pi y\)

Answer 27

okay yay!! :) so how do we do that last part?

200=πx + πy ?

Answer 28

yes ! solve y and plug that value in Area expression

Answer 29

200-πx=πy

y= 200-πx
-------------
π

?

Answer 30

yes, plug that value in Area expression

Answer 31

so x(200-πx/π) + (πx^2/4) + (π(200-πx/π)^2/4) ?

Answer 32

\(\large A = xy + \frac{\pi x^2}{4} + \frac{\pi y^2}{4}\)

\(\large A = x\frac{200 - \pi x}{\pi} + \frac{\pi x^2}{4} + \frac{\pi (200-\pi x)}{\pi^2 4}\)

Answer 33

yes, simplify it

Answer 34

and maximize

Answer 35

so

200x-πx^2 πx^2 200π-π^2x
---------- + ------- + -------------
π 4 π^2 4

?

Answer 36

\(\large A = xy + \frac{\pi x^2}{4} + \frac{\pi y^2}{4}\)

\(\large A = x\frac{200 - \pi x}{\pi} + \frac{\pi x^2}{4} + \frac{\pi (200-\pi x)}{\pi^2 4}\)


\(\large A = \frac{200x }{\pi} - x^2 + \frac{\pi x^2}{4} + \frac{50}{\pi } - \frac{x}{4}\)

Answer 37

^^

Answer 38

Maximize !

Answer 39

ahh not sure how to do this part :(

derivative then set to 0 right? :/ not sure what the derivative is though :(

Answer 40

yes, pi is just a constant.
dont get scared... dive in and find out the derivative of A

Answer 41

200 - 2x + 2πx - 1 ?

Answer 42

we made a mistake

Answer 43

oh no!! :(

Answer 44

Corrected below :

\(\large A = xy + \frac{\pi x^2}{4} + \frac{\pi y^2}{4}\)

\(\large A = x\frac{200 - \pi x}{\pi} + \frac{\pi x^2}{4} + \frac{\pi (200-\pi x)^{\color{red}{2}}}{\pi^2 4}\)

Answer 45

simplifying that gives :


\(\large A = \frac{200 x}{\pi} - x^2+ \frac{\pi x^2}{4} + \frac{ (200-\pi x)^{\color{red}{2}}}{ 4\pi}\)

Answer 46

Maximize this^

Answer 47

so 200x - 2x + 2πx + 400-2πx ?

Answer 48

nope

Answer 49

aww :(

Answer 50

\(\large A = \frac{200 x}{\pi} - x^2+ \frac{\pi x^2}{4} + \frac{ (200-\pi x)^{\color{red}{2}}}{ 4\pi}\)


\(\large A' = \frac{200 }{\pi} - 2x + \frac{\pi x}{2} + \frac{ 2(200-\pi x)(-\pi)}{ 4\pi}\)

Answer 51

set it equal to 0 and sovle x,

Answer 52

if u are lucky enough, u wud get :
http://www.wolframalpha.com/input/?i=maximize+xy+%2B+++%5Cfrac%7B%5Cpi+x%5E2%7D%7B4%7D+%2B+++%5Cfrac%7B%5Cpi+y%5E2%7D%7B4%7D%2C+%5Cpi*%28x%2By%29%3D200%2C+x%3E%3D0%2C+y%3E%3D0

Answer 53

ahh okay, yes i got that:)

thank you!!! :D

Answer 54

np :)

Answer 55

new problem:)

Answer 56

\(\large A = \pi r^2 + 2\pi r h\)

\(\large V = \pi r^2 h = 44\)

Answer 57

solve \(h\) from second equation and plugit in Area expression

Answer 58

h= 44/pi r^2 ?

Answer 59

A = pi r ^2 + 2 pi r (44/pi r^2 )

= πr^2 + 88
----
r

??

Answer 60

yup!

\(\large A = \pi r^2 + 2\pi r \frac{44}{\pi r^2}\)


\(\large A = \pi r^2 + \frac{88}{r}\)

Answer 61

Minimize this

Answer 62

2πr - (88/r^2) = 0

r= cube root 44/π ?

Answer 63

Excellent !!

Answer 64

yay! so for h, we plug that into which equation ? V =.... ?

Answer 65

find out \(h\) also :

\(\large \pi r^2 h = 44\)

Answer 66

yes^^

Answer 67

h= 44/πr^2

Answer 68

?

Answer 69

yes, plugin r value and simplify

Answer 70

not sure how to simplify h ;/

Answer 71

\(\large h = \frac{44}{\pi r^2} \)


\(\large h = \frac{44}{\pi \left([\frac{44}{\pi}]^{\frac{1}{3}}\right)^2} \)


\(\large h = \frac{44}{\pi \left(\frac{44}{\pi}\right)^{\frac{2}{3}}} \)



\(\large h = \left(\frac{44}{\pi}\right)^{\frac{1}{3}} \)

Answer 72

ahh okay thank you!! :D

Answer 73

which is same as \(r\)

Answer 74

So, the dimensions that minimize the surface area are :

\(\large r = h = \left(\frac{44}{\pi}\right)^{\frac{1}{3}} \)

Answer 75

ahh okay, thanks!!!! :D

Answer 76

u wlc :)