# Math Help - Pretty Basic Function Word Problems..

1. ## Pretty Basic Function Word Problems..

For this first one I tried to find a formula that I could relate to volume and surface area. Needless to say I didn't get very far..
I pretty much got stuck at the Surface Area Formula for rectangles: 2lw + 2lh + 2wh.
Help would be awesome.

I was pretty unsure of where to start for this one. I tried some stuff but got nowhere.

I'd really appreciate a helping hand, I have my first test coming up and I can't grasp this stuff.
Word problems have always been my Achilles heel.

2. If you have a square based box, its length and width are equal - call this length $x$. Call the height $h$.

Then $S = 2x^2 + 4xh$ and $V = x^2h$.

You're told that the Volume is $60\,\textrm{ft}^3$.

So $x^2h = 60$

$h = \frac{60}{x^2}$.

Therefore the surface area can be written as

$S = 2x^2 + 4\left(\frac{60}{x^2}\right)x$

$= 2x^2 + \frac{240}{x}$.

3. In order to write a function for the area of a rectangle, you need its length and width. So far you only have the width ( $h$).

If you split the rectangle in half, you will have two smaller rectangles, with width = $h$ and diagonal = $10$ (the radius of the circle is the same length as the diagonal).

So by Pythagoras:

$a^2 + b^2 = c^2$

$a^2 + h^2 = 10^2$

$a^2 + h^2 = 100$

$a^2 = 100 - h^2$

$a = \sqrt{100 - h^2}$.

If you double this, you have the length of the rectangle.

So $A = 2\sqrt{100 - h^2}\cdot h$

$A = 2h\sqrt{100 - h^2}$.

4. Thanks a lot for the help.
I understand everything.
One thing though, on the first problem.
How did we arrive at these formulas?
and .
I searched around the web for volume and surface area formulas like these but didn't discover any.

Also I was a bit stuck on this one too, if you or somebody could help i'd be thankful!
The part on the top right says "fence a garden plot"

5. Originally Posted by topdnbass
Thanks a lot for the help.
I understand everything.
One thing though, on the first problem.
How did we arrive at these formulas?
and .
I searched around the web for volume and surface area formulas like these but didn't discover any.
Try drawing the net of this box.

You should have two rectangles of dimensions $x \times x$ and four rectangles of dimensions $x \times h$.

So what is its surface area?

For the volume, we have a prism.

The Volume of a prism is $AD$, where $A$ is the cross-sectional area (in this case, the square base) and $D$ is the depth (or height).

Call the width of your rectangle $y\,\textrm{ft}$ (or any other letter).

So the cost is

$C = 5x + 3(x + 2y)$.

Can you see why?

You are also told that the area is $1200\,\textrm{ft}^2$.

So $xy = 1200$

$y = \frac{1200}{x}$.

Substituting back into the cost equation you get

$C = 5x + 3\left[x + 2\left(\frac{1200}{x}\right)\right]$

$= 5x + 3\left(x + \frac{2400}{x}\right)$

$= 5x + 3x + \frac{7200}{x}$

$= 8x + \frac{7200}{x}$.

To minimise the cost, differentiate the cost function, set equal to 0 and solve for $x$.

$C = 8x + 7200x^{-1}$

$\frac{dC}{dx} = 8 - 7200x^{-2}$

$= 8 - \frac{7200}{x^2}$.

Setting equal to 0:

$0 = 8 - \frac{7200}{x^2}$

$\frac{7200}{x^2} = 8$

$\frac{x^2}{7200} = \frac{1}{8}$

$x^2 = \frac{7200}{8}$

$x^2 = 900$

$x = 30$ (we can disregard the negative answer because you can't have a negative length).

To check that this IS a minimum, differentiate again, and check that it is positive at $x = 30$.

$\frac{d^2C}{dx^2} = 14400x^{-3} = \frac{14400}{x^3}$.

Clearly this will be positive for any positive $x$. So you do indeed have a minimum at $x = 30$.

So the dimensions are:

$x = 30, y = \frac{1200}{30} = 40$.

Now substitute these values into the cost function and see if you can afford it.

6. You are a saint, Prove It.

For the first one. I don't see where 4 rectangles (x*h) and 2 (x*x) come from.

Isn't this how the diagram should look?

Theres a rectangular box with a volume of 60ft^3. Below it is a square base.
We set the sides of the square base to X since they are the same.
If we split the rectangle (as I did in the diagram with a thin brush) we can break it up into smaller pieces, but why 4?
I think i'm pretty confused about the diagram and where all the rectangles come from.
Reading the problem implies that there is only a rectangle and a square base, as my diagram shows.
Where do the
two rectangles of dimensions and four rectangles of dimensions .
come from

Call the width of your rectangle y{ft} (or any other letter).

So the cost is

C = 5x + 3(x + 2y).

Can you see why?
Ok for this part on the 3rd problem, it's 5x (the cost per foot of fencing adjacent to the street (x)) then + 3(x + 2y). This specifies that the 3x is conditional (every side that's not next to the street) but I don't see exactly how.
It would make sense that the 2y is cause we have two y sides (width) that need to be accounted for.

7. Originally Posted by topdnbass
You are a saint, Prove It.

For the first one. I don't see where 4 rectangles (x*h) and 2 (x*x) come from.

Isn't this how the diagram should look?

Theres a rectangular box with a volume of 60ft^3. Below it is a square base.
We set the sides of the square base to X since they are the same.
If we split the rectangle (as I did in the diagram with a thin brush) we can break it up into smaller pieces, but why 4?
I think i'm pretty confused about the diagram and where all the rectangles come from.
Reading the problem implies that there is only a rectangle and a square base, as my diagram shows.
Where do the come from

Ok for this part on the 3rd problem, it's 5x (the cost per foot of fencing adjacent to the street (x)) then + 3(x + 2y). This specifies that the 3x is conditional (every side that's not next to the street) but I don't see exactly how.
It would make sense that the 2y is cause we have two y sides (width) that need to be accounted for.
A box has 6 faces... The top and bottom are $x \times x$, and the other 4 are $x \times h$...

For the second part, the $3(x + 2y)$ comes from the fact that the remaining fencing, $x + 2y$ costs \$3 per foot.