Question

world problem calculus

Answer 1

you are asked to design a retangular container with a square base and no top. your budget allows you to use a total of 300ft^2. of material for the sides and bottom. what are the dimensions of the largest capacity container you can design with this constraint

Answer 2

xy+2yz+2xz=300
xyz=V
now express z as function of x and y.
xy+z(2x+2y)=300
so z=(300-xy)/(2x+2y)
substitute it into Volume equation
xy(300-xy)/(2x+2y)=V
now find partial derivatives and set them to 0 to find posible critical points:
if need help to continue tell me

Answer 3

ty for helping first off where did u get the xy+2yz+2xz= 300? is that just the formula to know for a box?

Answer 4

or a rectangluar container

Answer 5

it is the sum of áreas of each side and bottom

Answer 6

bottom área is xy, right side área is zy, left side área is zy, front side área is xz and back side área is xz. Later just add it all and set it to máximum área alowed

Answer 7

yeah i get it now ok makes sense, i think I am going to have you continue on idk partial derivatives ?

Answer 8

is that just taking the volumne derivative?

Answer 9

nvm idont think what i said make any sense lmao....i dont know word problems at all sorry >_<!!!!!

Answer 10

i suck so much at these stuffs

Answer 11

anyone :[ can branch off what myko said ?

Answer 12

yes. For example the first derivative would go like this:
first rearange a bit the Volume expretion
V=300xy/2(x+y)-x^2y^2/2(x+y)
Now take derivative respect to x
V_x=[300y(2(x+y))-600xy/4(x+y)^2]-[4xy^2(x+y)-2x^2y^2]/4(x+y)^2=
[600xy+600y^2-600xy/4(x+y)^2]-{4x^2y^2+4xy^3-2x^2y^2]/4(x+y)^2=
[600y^2-2x^2y^2-4xy^3]/4(x+y)^2=[300y^2-x^2y^2-2xy^3]/2(x+y)^2=0
For this to equal 0, 300y^2-x^2y^2-2xy^3 have to be =0, because (x+y) not equal 0.
300y^2-x^2y^2-2xy^3=0
rearanging a bit:
y^2(300-x^2-2xy)=0
since y^2 not equal 0 , 300-x^2-2xy=0 or 2xy=x^2-300
this would be your first equation. Now do same for V_y

Answer 13

when you get 2º equation, solve the system for unknowns x and y and later substitute in área equation to find z. That's it

Answer 14

holy i read that like 5 times i might just give up on this problem move on thanks man xDD

Answer 15

im sorry im just slow >_< !!

Answer 16

its so much work for no reason omg lol

Answer 17

Sry I made sign mistake in last step:
should be: 2xy=300-x^2
when you do same for V_y you should get something like this:
2xy = 300 - y^2
now solve this 2 equations:
2xy=300-x^2 and 2xy = 300 - y^2
Making them equal:
300-x^2=300-y^2
this means x=y and also you know that x>0 and y>0 so:
substitute into V equation V=300xy/2(x+y)-x^2y^2/2(x+y)
you get:
V=300x^2-x^4/4x=300x-x^3/4
take derivative respect to x and set it to 0:
300/4-3x^2/4=0 or 75-3/4x^2=0 or 3/4x^2=75 or 3x^2=300 or x^2=100 or x=10, since you know it is x>0. Now just find the rest substituting this x value into some equation. For example, since you know x=y, y=10. Now just z left.

Answer 18

you can find it form here, xy+2yz+2xz=300
putting x=10 and y=10 you get:
100+20z+20z=300
this I think you cand do it alone, :)

Answer 19

yeah 40z = 200

Answer 20

but the steps to get here is too much for no reason

Answer 21

ima just skip the problem on the exam ima spend my time somewhere else now thnx so much

Answer 22

there is a more direct way, but you need LaGrange multiplyers for that

Answer 23

this way is more intuitive

Answer 24

steps are easy. Just invloves many calculations, that's it

Answer 25

this is highschool math

Answer 26

>_< !!

Answer 27

i really really hate word problems sigh lol

Answer 28

thats why i never paid attention in class and now im like fked Q_Q