Question

Optimization problem. Please help!

Answer 1

Something like that

Answer 2

In any optimization problem it is key to fabricate a function that determines the pattern of behaviour of the variables.
Since it's only the plane and we got a line belonging to that plane, we can draw the data:
\[A(x,y)=x.y\]
being the area of the rectangle, depending on two variables x,y.
\[y=\frac{ 2 }{ 3 }(8-x)\]
being the boundry line, now, since the boundry line is a geometric body belonging to the designated problem-area, we will just place it on the area formula in order to transform the function that represents the pattern of the area:
\[A(x,y) \rightarrow A(x) \iff A(x)=x.(\frac{ 2 }{ 3 }(8-x))\]
Giving as a result the function:
\[A(x)=x(\frac{ 16 }{ 3 }-\frac{ 2x }{ 3 })\]
\[A(x)=\frac{ 16x }{ 3 }-\frac{ 2x^2 }{ 3 }\]
\[A(x)=\frac{ -2x^2+16x }{ 3 }\]
This final function is the one you should work with, I assume you can do the rest.

Answer 3

What does A(xy) =x.y equal? the Area of the square yes?

Answer 4

yes it's the area of the rectangle, since it is bounded by the x and y's I just multiply them.

Answer 5

1) Is given
2) Unknown: Area of Square = xy
3) The picture
4) Is this what you came up with?

Answer 6

or was that 5) ?

Answer 7

The last one is most adecuate for (5)

Answer 8

Right right. Okay :) so when you plugged y into the xy (area of square) was that considered part 4?

Answer 9

Pretty much, yes, all I did was creating an optimization equation by getting rid of one variables.
There are ways to work it out with two variables, but partial derivative is not at your level just yet.

Answer 10

okay! I think I got it o.o hopefully :) thanks!

Answer 11

..So now to find the sides of the square shall I set y = 0 or x=0 to find both values?

Answer 12

using the
a(x) = -[2x^2+16x]/3

Answer 13

@Astrophysics I see you are here already! [that dog in your pic always makes my day]

Answer 14

Haha hey

Answer 15

Hii. Okay, let me upload a pic of what I have so far o.o

Answer 16

Haha you just have what owlcoffee wrote, what's next?

Answer 17

we like it xP
Well now i'm meant to solve for x and y from that a(x) = -[2x^2+16x]/3 ?

Answer 18

Do I set x = 0?

Answer 19

Ok so you have your function with just one variable right, so now you need to maximize your function, assuming this as I don't even know what your problem is haha. So take the derivative then set it equal to 0 which will give you your x value.

Answer 20

The problem is attached in the original question :}
Gotcha gotcha.

Answer 21

Ok so let me ask you this, what's your interval?

Answer 22

for the square? I think it is [0,8]

Answer 23

so where A<0 right

Answer 24

Yeah

Answer 25

a'(x) = -[4x+16]/3

Answer 26

Wait, then x=-4?

Answer 27

brb

Answer 28

back!

Answer 29

oh x=4