For All You Math Wizards!

[I.am]

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...

Work it out, you might learn something from it
I haven't solved it either thinking you can atleast do this

If you didnt do this way, there is no other way around.

-
I.am

[BiG_aL]

I don't quite understand your subsitution in it though.. it stumps me..

[I.am]

Originally posted by BiG_aL@15 February 2004 - 10:11
I don't quite understand your subsitution in it though.. it stumps me..

Work it out! Dont read it.

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

[Billy_Dean]

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.

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PS: Anyone who thinks l worked this out must be as silly as l am.

Thanx Da Wabbit.

[AussieSheila]

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!

Well done Da Wabbit!

[Billy_Dean]

Well if anyone should know me by now, you two should!