A farmer has 60m of fencing with which to construct three sides of a rectangular yard connected to an existing fence.
a) if the width of the paddock is x metres and the area inside the yard A square metres, write down the rule connecting A and x.

b) Sketch the graph of A against x.

c) Determine the maximum area that can be formed for the yard.

I didn't exactly understand this, and please keep it to year 11 standards. And I haven't learned calculas yet.

2. Originally Posted by Awsom Guy
A farmer has 60m of fencing with which to construct three sides of a rectangular yard connected to anb existing fence.
a) if the width of the paddock is x metres and the area inside the yard A square metres, write down the rule connecting A and x.

b) Sketch the graph of A against x.

c) Determine the maximum area that can be formed for the yard.

I didn't exactly understand this, and please keep it to year 11 standards. And I haven't learned calculas yet.
Year 11, isn't this a bit early for you to be doing this?

If the width of the paddock is x, this means that the 4 lengths will b x, x, 30-x and 30-x.

I'm sure you know that the area is equal to the width times the length, this gives us the equation $\displaystyle A = x(30-x) \rightarrow A = 30x-x^2$.

For part B, imagine that the $\displaystyle A$ was a $\displaystyle y$ and just sketch how you normally would.

As for the last bit, the best method for this would be calculus, but if you've not learnt it yet then I guess not in this case.

Here's a fact that should help you. For any shape with a set perimeter, the one that will give you the greatest area is the one with all sides equal.

For example, say you have 60 metres of fencing and you want to construct a triangular fence, the shape that will give you the greatest area would be the equilateral triangle, ie the one with all 3 sides 20m.

So if your building a rectangular yard, then the optimum shape would be...?

3. You were close but the answer that is on my sheet is A=60x-2x^2.

Originally Posted by craig
Year 11, isn't this a bit early for you to be doing this?

If the width of the paddock is x, this means that the 4 lengths will b x, x, 30-x and 30-x.

I'm sure you know that the area is equal to the width times the length, this gives us the equation $\displaystyle A = x(30-x) \rightarrow A = 30x-x^2$.

For part B, imagine that the $\displaystyle A$ was a $\displaystyle y$ and just sketch how you normally would.

As for the last bit, the best method for this would be calculus, but if you've not learnt it yet then I guess not in this case.

Here's a fact that should help you. For any shape with a set perimeter, the one that will give you the greatest area is the one with all sides equal.

For example, say you have 60 metres of fencing and you want to construct a triangular fence, the shape that will give you the greatest area would be the equilateral triangle, ie the one with all 3 sides 20m.

So if your building a rectangular yard, then the optimum shape would be...?

4. The fence that the 3 sides will be connected to, is that the width x ?

5. Originally Posted by craig
The fence that the 3 sides will be connected to, is that the width x ?
All I know is what I put up in the question.
Sorry about that, but thats all I too know.

Originally Posted by craig
The fence that the 3 sides will be connected to, is that the width x ?
also no we learn this stuff in year 11.

Originally Posted by craig
The fence that the 3 sides will be connected to, is that the width x ?
only two sides are the widths. That probably makes the 2x but how did they get 2x^2 (Confused)

Originally Posted by Awsom Guy
only two sides are the widths. That probably makes the 2x but how did they get 2x^2 (Confused)
It also says A square metres, since A can't be in in squared metres they have transfered it to the other side.

6. Originally Posted by Awsom Guy
It also says A square metres, since A can't be in in squared metres they have transfered it to the other side.
We make three sides of the paddock using $\displaystyle 60\ m$ of fencing. Then the area is $\displaystyle A=(60-2x)x$ (that is $\displaystyle x$ is the width of the paddock so we form two sides of length $\displaystyle x$ and the other is $\displaystyle 60-2x$), SO:

$\displaystyle A=-2x^2+60x$

We rewrite this (completing the square):

$\displaystyle A=-2(x^2-30x)=-2[ (x-15)^2+15^2]$

As $\displaystyle (x-15)^2 \ge 0$ for all $\displaystyle x$ the smallest value this can take is $\displaystyle 0$, when $\displaystyle x=15.$. But the minimum of $\displaystyle (x-15)^2$ corresponds to the maximum of $\displaystyle A$, so the maximum for $\displaystyle A$ is $\displaystyle 2\times 15^2$, and occurs when $\displaystyle x=15$.

CB

7. I don't need to find x I just need the answer A=60x-2x^2

Originally Posted by CaptainBlack
We make three sides of the paddock using $\displaystyle 60\ m$ of fencing. Then the area is $\displaystyle A=(60-2x)x$ (that is $\displaystyle x$ is the width of the paddock so we form two sides of length $\displaystyle x$ and the other is $\displaystyle 60-2x$), SO:

$\displaystyle A=-2x^2+60x$

We rewrite this (completing the square):

$\displaystyle A=-2(x^2-30x)=-2[ (x-15)^2+15^2]$

As $\displaystyle (x-15)^2 \ge 0$ for all $\displaystyle x$ the smallest value this can take is $\displaystyle 0$, when $\displaystyle x=15.$. But the minimum of $\displaystyle (x-15)^2$ corresponds to the maximum of $\displaystyle A$, so the maximum for $\displaystyle A$ is $\displaystyle 2\times 15^2$, and occurs when $\displaystyle x=15$.

CB

8. Originally Posted by Awsom Guy
I don't need to find x I just need the answer A=60x-2x^2