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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
http://www.chinkii.com/uploads/album...ath1234534.JPG
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
:huh:
i dont think thats possible
It was on a test, so it should be...Quote:
i dont think thats possible
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.
i read the question wrong
working........
thanks
I.am got it :frusty: :rolleyes:
nice work ;)
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 :P
http://www.sighost.us/members/Iam/area.gif
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
what did you get cuz we already had the test and it's tto late we jsut wanna know the answer...
Work it out, you might learn something from it :rolleyes:Quote:
Originally posted by BiG_aL@15 February 2004 - 10:08
what did you get cuz we already had the test and it's tto late we jsut wanna know the answer...
I haven't solved it either thinking you can atleast do this :D
If you didnt do this way, there is no other way around.
-
I.am
I don't quite understand your subsitution in it though.. it stumps me..
Work it out! Dont read it.Quote:
Originally posted by BiG_aL@15 February 2004 - 10:11
I don't quite understand your subsitution in it though.. it stumps me..
For the area part, its a rectangle and a square. Area of a rectangle is lxb and area of a square is side^2
Solving for perimeter we got the value of x in terms of y which is x=150-y
so Area A = 4(y)(y) + (150-y)^2
as A = lxb + side^2
I expanded (150-y)^2 in the equation.
I have a habit of skipping steps although I did minimum this time
Hope this helps,
I.am
I calculate both the maximum area and the minimum
areas possible.
Perimeter P = 300 = 2x + 10y. Hence, x = 150 - 5y.
The constraint that x be at least zero forces y to be at most 30m,
while the constraint that x is at most 4y forces y to be at least 150/9 = 50/3 ~16.667m.
Hence, any y which solves either the maximum or minimum area
problem is constrained to lie in the interval from 50/3 to 30m.
Now, we have the area of the figure is given by:
A = 4y² + (150 - 5y)².
Minimum area may occur when the derivative dA/dy = 0
(or at one of the extremes on y given above);
i.e., when 8y - 10(150 - 5y) = 58y - 1500 = 0.
So, y = 1500/58 = 750/29 ~ 25.862m
(which lies in the allowable interval of values given by the constraints above)
and x = 150 - 5(750/29) = 600/29 ~ 20.690m.
This yields an area (which, in fact, is a minimum by the 2nd
derivative test d²A/dy² = 58 > 0 ):
A = 4(750/29)² + (600/29)² = 2610000/841 ~ 3103.448m².
The only other possible extreme values for the area occur
when y takes on one of the extremes of its constrained range:
When y = 50/3 ~ 16.667m, we have
x = 150 - 5(50/3) = 200/3 ~ 66.667m.
Now, in fact x = 4y, so the figure is actually a rectangle of
dimensions: base = 4y = x = 200/3 ~ 66.667m
by height = y + x = 250/3 = 83.333m,
and the area for these dimensions is
A = (200/3)(250/3) = 50000/9 ~ 5555.556m².
At the other extreme of its constraints,
when y = 30m we have x = 0m,
and the figure is actually a rectangle with
base = 4y = 120m
height = y = 30m,
and the area for these
dimensions is
A = 120(30) = 3600m².
To summarize:
The minimum possible area is 2610000/841 ~ 3103.448m²
which occurs when x =600/29 ~ 20.690m
and y = 750/29 ~ 25.862m.
The maximum possible area is 50000/9 ~ 5555.556m²
which occurs when x = 200/3 ~ 66.667m
and y = 50/3 ~ 16.667m.
Nice problem. :)
--------------------------------------------------------------------------------------------
PS: Anyone who thinks l worked this out must be as silly as l am.
Thanx Da Wabbit.
:)
:lol: :lol: :lol:Quote:
Originally posted by Billy_Dean@16 February 2004 - 15:59
PS: Anyone who thinks l worked this out must be as silly as l am.
Thanx Da Wabbit.
:)
LMFAO! You had me going so much I messaged Cowsy halfway through!
:rolleyes: Well done Da Wabbit!
:)
Well if anyone should know me by now, you two should! :lol:
:)