Stuck on functions, please help.

G

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Hi.

I'm stuck on on the last part of this problem. Any help would
be greatly appreciated....I have answered parts 1 through
3, i believe correctly, now i just need help with 4 and 5.
thanks in advance.


Suppose you wanted to construct a fence around a garden plot in the form of a rectangle. On the neighbor’s side it’s going to need heavy-duty fencing that costs $2.00 per foot. The other three sides can be made of standard fencing material that costs $1.20 foot. You have $200 to spend.
Question 1 (10 points)
How many different rectangles would it be possible to enclose for $200?

It would be possible to enclose an infinite amount of rectangles as there are no size limitations set forth in the initial problem.




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Question 2 (20 points)
Write an equation using two variables for the total cost of the fence. Use x and y to represent the width and length of the rectangle. Solve the equation for one of the variables (either x or y).

2y + 1.2x + 1.2x + 1.2y = 200
2.4x + 3.2y = 200
3.2y = 200 - 2.4x
y = 200- 2.4x / 3.2
y = 50 - .6x / .8
y = 25 - .3x/.4
y = 250 - 3x/4

Save answer
Question 3 (20 points)
Given that Area = (x)(y), now write the area function for this problem using ONLY ONE VARIABLE (by making a substitution using the equation from #2).

Area = (x)(250-3x)/4

y = 250x - 3x^2 / 4


Save answer
Question 4 (40 points)
What dimensions will give the maximum area inside the fence?

???

Save answer
Question 5 (10 points)
What will be the maximum area?

???
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bqdla said:
Write an equation using two variables for the total cost of the fence. Use x and y to represent the width and length of the rectangle. Solve the equation for one of the variables (either x or y).

2y + 1.2x + 1.2x + 1.2y = 200
2.4x + 3.2y = 200
3.2y = 200 - 2.4x
y = 200- 2.4x / 3.2
y = 50 - .6x / .8
y = 25 - .3x/.4
y = 250 - 3x/4
I'm going to pick on this one a little:

First, I think you should start with "less than or equal to", since the problem statement does not state that you must spend the ENTIRE sum. Perhaps this is an unnecessary complication, but based on the anser to #1, perhaps it is appropriate.

Second, how did you do this?

y = 200- 2.4x / 3.2
y = 50 - .6x / .8

That would be tons more clear with the addition of some parentheses:

y = (200 - 2.4x) / 3.2
...
y = (250 - 3x)/4
 
Hi,

Thanks for responding. You're correct. I should not have used the = sign, but it should have been like this:

2y + 1.2x + 1.2x + 1.2y <= 200
2.4x + 3.2y <= 200
3.2y <= -2.4x +200
y >= (-2.4x + 200)/3.2
y >= (-.3x + 25)/.4 = (-3x + 250)/4

I added the parentheses as well.....now, for question #4 or #5 :)

Help is greatly appreciated

Brian
 
bqdla said:
What dimensions will give the maximum area inside the fence?

It's probably obvious that the maximum area will be achieved by using the entire $200. This allows us to abandon the "less than" part.

Hint: Area Inside Fence = x*y = x*¼*(250-3*x)

See if that makes any sense. Then, what do you do with it?
 
I'd set x*y =200 and then solve for each one? is that right?
 
tkhunny said:
Hint: Area Inside Fence = x*y = x*¼*(250-3*x)
Use the hint, Luke...

Unfortunately, we are just a bit confused, due to our notation. One side of the fence is labled 'y'. This was an unfortunate choice. We'll have to use function notation in order not to confuse ourselves.

If you encountered f(x) = x*¼*(250-3*x), and were instructed to find its maximum value, what would you do?
Do you recognize the form? (Parabola, quadratic)
Is this the first time you ever faced such a problem?
Can you find where f(x<sub>0</sub>) = 0 (factoring, completing the square, quadratic formula -- Of course, it's already factored, so that might be the best way to go.)

Always wishing to stress Domain considerations, one probably should narrow the search for answers. In this problem, since we are building a real fence, we have y > 0 and x > 0. How about the other direction? If we build the entire fence parallel to the neighbors, we have 3.2*y <= 200, or y <= 62.5. Similarly, if we build a very narrow fence perpendicular to the neighbors, we have 2.4*x <= 200, or x <= 83.333.... This helps often. If you get x = -5, throw it out. If you get y = 73, throw it out. They just are not in the Domain.
 
bqdla said:
3.2y <= -2.4x +200
y >= (-2.4x + 200)/3.2
By the way, why did you switch the direction of the inequality? Dividing by positive numbers doesn't do that.
 
ok, so i think function notation is the culprit, its always been the culprit with me in algebra.....so what i need to do is not think of it as y = something, but as f(x) = (x(250-3x))/4 ....as far as the maximum value is concerned, that was my original question, so i don't really know. In class, every time the book mentions max or min or whatever, its done with a graphing calc, so I'm not sure how to do it by hand :(


it would be a quadratic, i think,
 
and i flipped the inequality by mistake, i forgot it was only when using negatives....
 
bqdla said:
f(x) = (x(250-3x))/4
In class, every time the book mentions max or min or whatever, its done with a graphing calc
it would be a quadratic
Oh, Dear, the curse of the calculator.

It is quadratic. You can find where it is zero using any available methodology. In this case, it's already factored. Let's go with that. (There's another cool trick if it isn't already factored, but I'll let that go for now.)

Where is f(x) = 0?

(x(250-3x))/4 = 0
x(250-3x) = 0

x = 0 or 250 - 3x = 0
x = 0 or x = 250/3

There's a secret, here. The min or max for one of these quadrtic things is ALWAYS right in the middle of the two zeroes. Calculate the arithmetic mean (the average).

x<sub>min/max</sub> = (250/3 + 0)/2 = 250/6 = 125/3

f(125/3) = ??? I'll let you do that.

You can verify the results on your graphing calculator. You HAVE to understand what you are doing. A calculator never will be a good substitute for understanding.
 
For the record, I agree with you about the calculators. i've just gone back to school after almost 15 years since the first time, and I'm kind of shocked just how much they are used. My first go around in high school and college, we weren't even allowed to use calculators....ever! But i digress.....

so anyway, thanks for that on the max and min, its cleared up a lot of things for me...so

f(125/3) = ((125/3)*(250-3)*(125/3)) / 4 = 107204.86

is that right? how does that translate to the maximum?
 
At some point, you have to remember what it was we were doing. :)

x is a distance, in feet.

x = 125/3 means 41 + 2/3 ft

Thanks to your earlier work, this also gives 'y'

y = ¼*(250 - 3*x) = ¼(250 - 3*(125/3)) = ¼(250 - 125) = ¼*(125)

y = 125/4 means 31¼ ft

The total area, for thus x-y combination is x*y = (125/3) * (125/4) = 15625/12

x*y = 150625/12 means 1302.08333... ft<sup>2</sup>

You may also wish to check how you substituted your x-value. That didn't work very well.

f(x) =¼*x*(250-3*x) = ¼*(125/3)*(250-3*(125/3)) = ¼*(125/3)*(250-125) =

You try it from there.
 
Thank you so much for all your help and patience....I'm all set now, thanks again!!
 
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