# Thread: Word problems, For Functions

1. ## Word problems, For Functions

Is it possible to build a fence on 3 sides of a rectangular piece of land with an area of 100m^2, so that the total length of the fence is 25 m ?

I created the two equations l+2w=25m (same as l=25-2w) and l*w=100m^2 . Then I put them together to make (25-2w)(w)=100 , and then I distribute the "w" and complete the square. The back of the book says it isn't possible though. Anyone able to do this and tell me what you get? Because I AM getting a final answer, but its not working out.

2. Originally Posted by mike_302
(25-2w)(w)=100
Please double check to see whether this quadratic equation has real solutions.

Roy

3. Originally Posted by mike_302
Is it possible to build a fence on 3 sides of a rectangular piece of land with an area of 100m^2, so that the total length of the fence is 25 m ?

I created the two equations l+2w=25m (same as l=25-2w) and l*w=100m^2 . Then I put them together to make (25-2w)(w)=100 , and then I distribute the "w" and complete the square. The back of the book says it isn't possible though. Anyone able to do this and tell me what you get? Because I AM getting a final answer, but its not working out.
Well lets think of this in a different way!

lets find the max area that can be enclosed with 25 meters of fence.

we know that

$\displaystyle A=l \cdot w \mbox{ and } l=25-2w$

subbing the 2nd equation into the first we get

$\displaystyle A=(25-2w)w$ if we complete the square on this we get

$\displaystyle A=-2 \left( w-\frac{25}{4}\right)^2+\frac{625}{8}$

$\displaystyle \frac{625}{8}=78.125$

So the maximum area that can be enclosed is 78.125 square feet.

4. thats the thing.. I'm getting the following:

-2w^2+25w=100
-2(w^2-25w+625/4-625/4)=100
-2(w-25/2)^2 + 625/2 = 100
....
....
....
where am i going wrong?

5. Should this

Originally Posted by mike_302
-2(w^2-25w+625/4-625/4)=100
be $\displaystyle -2\left(w^2-\frac{25}{2}w + \frac{625}{16}-\frac{625}{16}\right)=100$ ?

Roy

6. OH! Ok, now I see what I am not doing. When I see that 25/2 , I always assume that I already divided by 2, but i forget I have to divide by 2 again (mulitply 25/2 by 1/2) and THEN square.

Thanks! Took a bit of thinking to realize my error.

On a side note, I'm doing another question. I'm fairly confident that I can complete the square this time, no problems, but I can't figure out the equation to begin with!

The word problem:
The hypotenuse is 3 cm greater in length than the next longest side of a right triangle. The line perpendicular to THAT line is 3 cm shorter . What are the side lengths. (so, you have your hypotenusse, c, which is 3 cm longer than side b, which is 3 cm longer than side c )

7. Let c be the hypotenuse, let a be the longer leg, let b be the shorter leg.
Base on the given, we have a = c-3, b = a-3 = c-6, by Pythagorean theorem, we have
$\displaystyle c^2=(c-3)^2+(c-6)^2$
From here, you can find all 3 sides.

Roy