RE:
Go the following link => http://www.mathhelpforum.com/math-he...20-2007-a.html and see #4. Look around the forum as we have already solved millions of these optimization problems before.
-qbkr21
Im on some homework and came across this problem.
A contractor is to fence off a rectangular field along a straight river, the side along the river requiring no fence
a) What is the least amount of fencing needed to fence off 30,000m^2
I'm assuming im supposed to use 2X + Y = 30,000m^ for perimiter. However i know that i cant have two variables in the function. What'd be the function from here?
and b) What is the greatest area the contractor can fence off using 500m of fencing.
This is trying to find the maxima for area... and the formula would be (x^2)(y) for Volume yes? how do you get the function from here? Im pretty sure I can finish it after i get the function though
__________________________________________________ ___--
And theres another problem..
dont remember how its worded but it wants the minimum of the sum of two positive numbers, where the first is ^3 and the second is ^2. Also, they, without the ^, = 4
so
X+Y=4
Y= (-x+4)
X^3 + (-x+4)^2 -> X^3 + X^2 -8X +16
F'= 3x^2 + 2x -8
and after using -b plus/minus (Radical b^2 (-4)(a)c) divided by 2a
the x's are .. 1.33333333333etc and .. -2 i think
and the 1.3333333333333 comes to be the min because after plugging it into f" = 6x +2 it is a Positive concave,
so 1.3333333333 being x, i'd put it into f' to find the y, which comes to be.. around 9.45 or something similar,
and putting that into the origional equation, (1.33333333^3) + (9.45^2) = around 95 or a high number like that
HOWEVER this MUST be wrong because 9.45 + 1.3333 does NOT = 4
Where did i mess up on this one?
thanks for your help!
RE:
Go the following link => http://www.mathhelpforum.com/math-he...20-2007-a.html and see #4. Look around the forum as we have already solved millions of these optimization problems before.
-qbkr21
No. 2X + Y is the perimeter, or length, of the fence.
X*Y = 30000.
P = 2X + Y
You wish to minimize P. Thus:
Y = 30000/X ==> P = 2X + 30000/X
dP/dX = 2 - 30000/X^2 = 0
2 = 30000/X^2
X^2 = 30000/2 = 15000
X = 122.47 m
Is this a minimum point? Well,
d^2P/dX^2 = -60000/X^3 > 0 for X = 122.47
Thus this is indeed a minimum point. I'll let you decide if this is merely a relative minimum or an absolute minimum.
X = 122.47 m ==> Y = 30000/122.47 = 244.95 m
So
P = 2[122.47] + [244.95] = 489.90 m
-Dan
What the heck are you doing in the section in red?
And if they want the minimum sum of the two numbers how do you know it is 4?
Okay, assuming your setup is correct you want the minimum of the function
F = X^3 + X^2 - 8X +16
So
dF/dX = 3X^2 + 2X - 8 = 0
X = -4/3, 2
For which value of X is F a minimum?
Well,
d^2F/dX^2 = 6X + 2
For X = -4/3, d^2F/dX^2 = 6*(-4/3) + 2 = -6 < 0 so this represents a local maximum.
For X = 2, d^2F/dX^2 = 6*2 + 2 = 14 > 0 so this represents a local minimum. (Again I'll let you decide if this is an absolute minimum.)
So X = 2 ==> Y = -2 + 4 = 2, so the two numbers are 2 and 2.
-Dan
The X + Y = 4 was given in the question.And if they want the minimum sum of the two numbers how do you know it is 4?
what i want the min for, would be X^3 + X^4
for the part in red i didnt have the question with me, it was one i did earleir and was trying to remember it
however i think what i did wrong, was i messed up the negative and positive signs partway instead of -4/3, 2, i had 4/3, -2. <.<
gah ive been doing too much today =PVolume?? You're looking for area!