For All You Math Wizards!

[BiG_aL]

Aight me and a few other got this question the other day in math class and we can't figure it out.
you have this diagram



The garden has a perimeter of 300m. Find the value for X and a value for Y that produce a garden with a minimum area.


The sides with out variables need to be included in the 300 aswell. Can anyone explain how to do this question? Along with a answer?
Thanks,
Allan Melvin

[j4y3m]

[TARPD]

i dont think thats possible

[kaiweiler]

i dont think thats possible

It was on a test, so it should be...

[BiG_aL]

yeh it's in my math book and our physics teacher said it wasn't possible but our "subsitute" math teacher said it is...
I dunno I neec an answer.. There is a formula but I dunno what and how too look at it etc... it's fuckzored.

[TARPD]

i read the question wrong



working........

[BiG_aL]

thanks

[(>Zero Cool<)]

I.am got it

nice work

[I.am]

You dont need math wizards for a simple problem.

I wont solve the whole problem but most of it. The idea is that you know how to do the problem not just copy it



Assuming one side to be z, so other automatically becomes 4y-x-z

We know perimeter of this shape is 300.
So,
y+z+x+x+4y-x-z+y+4y=300
Solving we get, x = 150-5y

The shape can be divided into a rectangle and a square.
So area,

A = 4(y)(y) + x^2

Substituting the value of x from the previous equation, we get

A = 4y^2 + (150)^2 + (5y)^2 -2(150)(5y)

Dont bother expanding.
Now
Take the first derivative of A wrt y and put it equal to 0.
Find the value of y and take the second derivative of this equation. Plug the value of y in that equation and see if you get a positive answer. If its positive then function A has a minimum at that point.

Now from y get x using x=150-5y.
And then you are done.

Hope this helps,
I.am

[BiG_aL]

what did you get cuz we already had the test and it&#39;s tto late we jsut wanna know the answer...