2 Questions -Need Help Badly

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March 29th 2006, 06:18 AM
OverclockerR520

2 Questions -Need Help Badly

Hello nice to meet everyone this is my first post, I'm in need of help badly, regarding 2 precalc questions

Hello,

I have a math hand in due for class this evening around 5pm. There are several questions on this handout relating to trig and calc but these two problems I'm having trouble setting up the correct formulas and even creating the correct formulas to finish off the problem. Below are the copied and pasted questions, there were no diagrams given with the questions us our class to help, sorry.

Any help would be very much appreciated, thanks.
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2) An open box is to be made from a rectangular piece of cardboard having dimensions 20 inches by 30 inches by removing squares of area x2 from each corner and turning up the sides.

a) Show that there are two boxes that have a volume of 1000 cubic inches. Give their dimensions.

b) Which box has the smaller surface area?

3) A man wishes to put a fence around a rectangular field and then subdivide the field into three smaller rectangular plots by placing two fences parallel to one of the sides. If he can afford only 1000 yards of fencing, what dimensions will give the maximum rectangular area?

a) Express y as a function of the length x.

b) Express the total enclosed area A of the field as a function of x.

c) Find the dimensions that maximize the total enclosed area.
March 29th 2006, 12:55 PM
topsquark

Quote:

Originally Posted by OverclockerR520

2) An open box is to be made from a rectangular piece of cardboard having dimensions 20 inches by 30 inches by removing squares of area x2 from each corner and turning up the sides.

a) Show that there are two boxes that have a volume of 1000 cubic inches. Give their dimensions.

b) Which box has the smaller surface area?

2) The dimensions of the box are: l = 30 - 2x, w = 20 - 2x, h = x. So the volume of the box is: V = (30 - 2x)(20 - 2x)x.

So. The volume of the box is 1000.
$1000=(30 - 2x)(20 - 2x)x = 4x^3-100x^2+600x$

$4x^3-100x^2+600x-1000=0$

$x^3-25x^2+150x-250=0$

I don't know where you are as far as solving cubic equations is concerned. I will simply state at this point that the solutions are x = 5, x = $10+5\sqrt2$ , and x = $10-5\sqrt2$ . If you need help with this sort of thing, just say so and I'll be happy to provide.

So which are the two we can use for the box? Remember that we are removing 2x from each dimension of the cardboard. $30-2(10+5\sqrt2)=10-10\sqrt2<0$ , so we can't use that one. The other two will work.

So we have two possible boxes:
1) $l = 30 - 2*5 = 20$ in
$w = 20 - 2*5 = 10$ in
$h = x = 5$ in

2) $l = 30 - 2(10-5\sqrt2) = 10 + 10\sqrt2$ in
$w = 20 - 2(10-5\sqrt2) = 10\sqrt2$ in
$h = x=10-5\sqrt2$ in

As for surface area: We have at the bottom of each box a rectangle of area (30 - 2x)(20 - 2x). We have two vertical flaps of area (30 - 2x)x and two of area (20 - 2x)x. Adding all this up we get a surface area of:
$SA=(30-2x)(20-2x)+2(30-2x)x+2(20-2x)x$
$=-4x^2+600$

1) x = 5: SA = 500 in^2

2) $x = 10-5\sqrt2$ : $SA = 400\sqrt2$ in^2

So box 1 has the smaller surface area.

-Dan
March 29th 2006, 01:15 PM
topsquark

Quote:

Originally Posted by OverclockerR520

3) A man wishes to put a fence around a rectangular field and then subdivide the field into three smaller rectangular plots by placing two fences parallel to one of the sides. If he can afford only 1000 yards of fencing, what dimensions will give the maximum rectangular area?

a) Express y as a function of the length x.

b) Express the total enclosed area A of the field as a function of x.

c) Find the dimensions that maximize the total enclosed area.

Hard to do without a picture. Okay, I am going to use the following orientation: x is the horizontal dimension on the paper and y is the vertical dimension. I am assuming the two interior fences are vertically oriented.

a) The perimeter + length of the two interior fences = 1000 yd. So:
$1000 = 2x + 2y + 2y$
where the last "2y" comes from one "y" for each fence. (If you had the interior fences horizontally oriented then you get a "2x" here.)
So I get $y = \frac{1000-2x}{4}=250-\frac{x}{2}$

b) $A=xy=x \left ( 250-\frac{x}{2} \right ) = 250x - \frac{x^2}{2}$

c) I am assuming here, since you don't have Calculus at your beck and call that you would be using the properties of parabolas to figure this one out. Note that the area as a function of x is a parabola opening downward, so the vertex of the parabola gives the maximum area. The axis of symmetry of the parabola $A=-ax^2+bx+c$ is $x = \frac{b}{2a}$ . (Usually this formula has a negative sign in it. In this case I've explicitly made the coefficient of x^2 negative so it goes away.) Your parabola is:
$A=- \frac{x^2}{2}+250x$

So the line of symmetry will be $x=\frac{250}{2* \frac{1}{2}}=250$ .

So let x = 500 yd. in your area equation:
$A=-\frac{250^2}{2}+250(250) = 31250$ yd^2.

(Note: Different pictures of the fenced in area will give slightly different answers to a) and b) so you should include a diagram so your teacher can see what you are up to. The answer to c) will be the same regardless of your diagram.)

-Dan
April 3rd 2006, 03:54 AM
OverclockerR520

Thanks!

:) Wow,

Thanks everyone very much. Much more than I expected. Sorry for the delayed reply Ive been quite busy. But like I said this was a huge help, thanks again.

Brian