# I have some problemss in Calculus

• Sep 25th 2006, 07:21 AM
Hippocrates1985
I have some problemss in Calculus

Take-Home Problems for Test I Math 131A

Thank you.
• Sep 25th 2006, 07:25 AM
ThePerfectHacker
I am probably unathorized aid :eek:
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1)p(x)=4x and q(x)=x

2)p(x)=x and q(x)=x^2

3)p(x)=x^2 and q(x)=x

4)p(x)=x+2 and q(x)=x
• Sep 25th 2006, 07:31 AM
Hippocrates1985
Don't worry my friend "ThePerfectHacker." I'm not in that university but I like to visit that page for that professor to see his problems to his students.

I appreciate that from you.

Thank you.
• Sep 25th 2006, 07:31 AM
ThePerfectHacker
Complete the square picture, draw a dotted line to see a square and the equilateral on top.

Now, that dotted line has length x so the sides of the equilateral are too x. Also, that dotted line is a sides of a square thus the square has side of x.
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This is the part I do not understand. The question says the perimeter is 17 then adding the sides we have 5x=17 thus, x=3.4 But it seems to me that is not what the problem is saying. Why you placed it there I do not know.
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If x is the length of the sides of that pentagram then we have:
AREA OF EQUILATERAL=x^2*sqrt(3)/4
AREA OF SQUARE=x^2
TOTAL AREA=x^2+x^2*sqrt(3)/4

This is parabola is with a minimum, it has no upper bound, except at the endpoints which you fail to mention.
• Sep 25th 2006, 08:58 AM
CaptainBlack
Quote:

Originally Posted by Hippocrates1985
Don't worry my friend "ThePerfectHacker." I'm not in that university but I like to visit that page for that professor to see his problems to his students.

I appreciate that from you.

Thank you.

Since you are posting from an IP address that hails from Valpariso U,
as does the test, I think you own ImPerfectHacker an appology.

RonL
• Sep 25th 2006, 09:21 AM
ThePerfectHacker
I am going to send the professor an e-mail declaring that you cheated, so you can fail the course.
• Sep 25th 2006, 10:59 AM
Soroban
Hello, Hippocrates1985!

Quote:

2. A window has the shape of a rectangle surmounted by an equilateral triangle.
The perimeter is 17 feet.

(a) Express the area of the window as a function of the length of one side of the triangle.
Be sure to give the domain of the function.

(b) Submit either a print-out or a sketch of a graph of the function you found in part (a).

(c) Use the graph you made in part (b) to determine the side of the triangle
that gives the largest possible area of the window.

Code:

```          *         / \       x /  \ x       /    \       *      *       |      |     y|      |y       |      |       *-------*           x```

Let x = side of the triangle (and width of the rectangle)
Let y = length of the rectangle.

The perimeter is: .3x + 2y .= .17 . . y .= .(17 - 3x)/2 . [1]
. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . ._
The area of an equilateral triangle with side x is: (√3/4)x²
The area of the rectangle is: xy
. . . . . . . . . . . . . . . . . . ._
The total area is: .A .= .(√3/4)x² + xy
. . . . . . . . . . . . . . . . . _
Substitute [1]: . A .= .(√3/4)x² + x(17 - 3x)/2
. . . . . . . . . . . . . ._
Therefore: .A .= .(√3/4 - 3/2)x² + (17/2)x . (a)
. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . _
Since A must be nonnegative: . 0 .< .x .< .(34/33)(6 + √3)

(b), (c) If this is a Calculus problem, graphing is not necessary.
In fact, the graph gives us only an approximate answer.

The area function is a quadratic, hence its graph is a parabola
. . . . . . . . . . . . . . . . . . . . . . . . . . . . . _
. . with its vertex at: . x .= .(17/13)(6 + √3) . .3.983

"Eyeballing" the graph, the vertex seems to be at (4,17)
• Sep 25th 2006, 11:26 AM
Hippocrates1985
First of all,

My friend CaptainBlack, I am not a lier I said I am not in that university and that means I am not a student in that university. I am studing at MSU but I am here at Valparaiso Universty to visit my friend here. If you do not believe me I can enter from MSU tomorrow and you will see.

Second,

My friend ThePerfectHacker I did not ask you to give me the answer. I asked you to help me.

I am sorry if there is a miss understanding from you and in the same time thank you for your kindess.
• Sep 25th 2006, 12:43 PM
ThePerfectHacker
Quote:

Originally Posted by Soroban
Hello, Hippocrates1985!

Code:

```          *         / \       x /  \ x       /    \       *      *       |      |     y|      |y       |      |       *-------*           x```

Let x = side of the triangle (and width of the rectangle)
Let y = length of the rectangle.

The perimeter is: .3x + 2y .= .17 . . y .= .(17 - 3x)/2 . [1]
. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . ._
The area of an equilateral triangle with side x is: (√3/4)x²
The area of the rectangle is: xy
. . . . . . . . . . . . . . . . . . ._
The total area is: .A .= .(√3/4)x² + xy
. . . . . . . . . . . . . . . . . _
Substitute [1]: . A .= .(√3/4)x² + x(17 - 3x)/2
. . . . . . . . . . . . . ._
Therefore: .A .= .(√3/4 - 3/2)x² + (17/2)x . (a)
. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . _
Since A must be nonnegative: . 0 .< .x .< .(34/33)(6 + √3)

(b), (c) If this is a Calculus problem, graphing is not necessary.
In fact, the graph gives us only an approximate answer.

The area function is a quadratic, hence its graph is a parabola
. . . . . . . . . . . . . . . . . . . . . . . . . . . . . _
. . with its vertex at: . x .= .(17/13)(6 + √3) . .3.983

"Eyeballing" the graph, the vertex seems to be at (4,17)

I assumed that same thing but the problem is the problem say,
"equilateral" triangle. Meaning all sides of that figure must be the same.
• Sep 25th 2006, 02:01 PM
Hippocrates1985
Thank you guys for helping me. Actually I do not need the answers. I want to anderstand everything. For that I will give my answers and please correct me.

Actually I have the same answer to (b) and (c). For (a) and (c) I have these answers:-

a) q(x)=x^3 + 4x^2 +x and p(x)=x^2 + x

q(x)/p(x)

then by dividing the numerator and denominator by the highest power of x in the denominator.

c) q(x)=x^3 + x^2 + x and p(x)=x^2 + x

q(x)/p(x)

then by dividing the numerator and denominator by the highest power of x in the denominator.

I'm working now with problem #2 to understand it very well. Then, I will write my answer.

Thank you.
• Sep 25th 2006, 02:12 PM
Soroban
Hello, TPtHacker!

Quote:

I assumed that same thing but the problem is
the problem says: "equilateral" triangle.

which means that all sides of the triangle must be the same.

Otherwise, why did they refer to a rectangle instead of a square?

If the base must be a square, the problem is elementary geometry.

All sides will be 17 ÷ 5 = 3.4 units.
You can calculate the area of the triangle and the square: about 7.8956 square units.
. . Not what I call a Calculus problem . . .

• Sep 25th 2006, 02:51 PM
Hippocrates1985
My friend Soroban,

I cannot understand this part:-

Quote:

Substitute [1]: . A .= .(√3/4)x&#178; + x(17 - 3x)/2
. . . . . . . . . . . . . ._
Therefore: .A .= .(√3/4 - 3/2)x&#178; + (17/2)x . (a)
. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . _
Since A must be nonnegative: . 0 .< .x .< .(34/33)(6 + √3)
What happend to x which located with 3 in x(17-3x)/2 ?
• Sep 30th 2006, 01:54 AM
classicstrings
Seems quite a coincidence that this is "Due September 26, 2006".

Oh and the advertisement "g a y wedding rings"... this is a hoot, it is censored when I write ***, but the advertisement is allowed....
• Sep 30th 2006, 02:30 AM
CaptainBlack
Quote:

Originally Posted by classicstrings
Seems quite a coincidence that this is "Due September 26, 2006".

Oh and the advertisement "g a y wedding rings"... this is a hoot, it is censored when I write ***, but the advertisement is allowed....

I thrust that you have seen that our "friend" claims that he is not a
student at the institution that the take home test hails from.

How this quite goes with his never having posted from anywhere else
(I've checked his IP address/s) I don't know, but he claims he has a
explanation for that as well.

Does this make us look stupid - yes.

RonL