CPhill

Hannah has 2000 feet of fence. She is going to make a rectangular garden using the fence and her barn for one side (so she only needs fencing on three sides of the garden). What is the largest area she can enclose? [Answer with just the number of square feet of area].

Let  two  equal sides of the fence  =  S

Then the remaining side  (opposite the barn)   is   just  2000  - 2S

So....the area, A,   is just

A  = S (2000 - 2S)    simplify

A  = 2000S  - 2S^2         take the derivative of this and set to 0

A'  = 2000 - 4S  = 0

Solving, we have

2000  - 4S  = 0      add 4S  to both sides

2000  = 4S   divide both sides by 4

500  = S

So....the max area  is  S  (2000 - 2S)   =      500 ( 2000 - 2*500)   =   500 (1000)  =   500,000  ft^2

Note that this just boils down to counting the arrangements  of  sets

To get from A to the intermediate point  we can go

(East, East, North, North, North)

And  the total number of arrangements of this set  is  just place East  in any of the two of the  five positions  (or, alternatively, North in any of three of the five positions)...so  C(5,2)   = C (5,3)   = 10  ways

Next...to get from the intermediate point to B we have

(East, East, East , East, North , North , North)

So....the possible arangements of this set  is either placing East  in any four of the seven positions or to place North in any of three of the seven positions  =   C( 7,4)  =  C(7,3)   =   35  ways

So...the number of ways to  get from A to B  =   10 * 35   = 350  ways

Here's  4

ax^2   + 8x  +  4   = 0 

If this has only one solution.....then it will be true  that the  discriminant given by  (8)^2  - 4 (a) (4)  will = 0

So we have

(8)^2  - 4 (a) (4)   =  0   simplify

64  - 16a    = 0      add   16a  to both sides

64   = 16a        divide both sides by 16

4  =  a 

Here's  (1)   without a calculator....note that the numerator and  denomonator  are  both just the difference of squares...so we have :

(2011 + 2006) ( 2011  - 2006)       

_______________________    =

(2010 + 2007) ( 2010 - 2007)

(4017) (5)

_________  =

(4017) (3)

 5

__

 3

a)

x + y  = -1

3x  = 4  - 3y

Rearrange the first equation as    y  = -1  - x

Sub this into  the second equation for  y

3x  = 4  - 3(-1 - x)   simplify

3x = 4 + 3 + 3x

3x  = 7 + 3x       dubtract   3x from both sides

0  = 7      this is absurd....so....this system has no solution

b)

3x - 4y  + 2   = 0  ⇒     3x  -  4y  = -2       (1)

10 - 10y  = 10y - 15x

Divide the second equation through by   5 and we have

2 - 2y  = 2y   - 3x       rearrange as

3x - 4y  = -2       (2)

Notice that equations 1  and 2 are exactly the same.....so....when this happens....we have  infinite solutions for both  x and  y

If we cut  2 inch squares form the corners, the dimensions  of the box will be

Length  = ( x - 4) 

Width  = ( x + 2  - 4)    =  ( x  - 2)

Height   = x

So

Length  * Width  * Height  =  96

(x - 4) ( x - 2)  (x)   = 96      simplify

(x^2 - 6x + 8)(x)  = 96

x^3 - 6x^2 + 8x  = 96

x^3 - 6x^2 +  8x  - 96  = 0  

We can solve this with a graph  :

https://www.desmos.com/calculator/iopnxaxlca

The x value that make the above function  =  0  is ≈   6.87  in

So....the  approximate dimensions  of the box  (in inches) are

(6.87 - 4)  ( 6.87  - 2) (6.87)  =

(2.87) ( 4.87) (6.87)

The volume of the box  is  L * W * H.....but..... L and  W are the same  = the side lengths

So...let the side lenghts  = S

And  the height is   S  + 7

So...we have that

S * S * (S + 7)   = 4000       simplify

S^3  + 7S^2   = 4000

S^3  + 7S^2  -  4000   = 0

We can obtain a graphical solution here :

https://www.desmos.com/calculator/51gbh74ttk

The graph shows that the side length  ≈  13.85  in

But the height  is 7 more  ≈  20.85  in

\(q(x) = \frac{\sqrt{x}}{\sqrt{1-x^2}}\)

Note that.....for the function in the numerator, the domain  is     [ 0, ∞ )

For the function inthe denominator, the domain is (-1, 1)

But....we must use the most restrictive domain possible.....this is  [ 0, 1 )

Here is a graph that will confirm  this  :

https://www.desmos.com/calculator/ywedl27com

I believe we  need  to include 60, as well

So...the sum  is   66 + 60   =   126

Let  the  number of rows of the original formation  =  R

And let the  number of columns  =  C

So   R * C  = 105   ⇒    C  = 105/R       (1)    

Now...in the new formation.....the  number of rows  is  ( R + 6)  and the number of columns = (C - 2)

So we have that   

(R + 6) (C - 2)  = 105  ....  expand...   

R*C + 6C - 2R - 12  = 105   

105 + 6C - 2R  - 12 = 105     ...simplify...

6C - 2R =  12      sub  (1)  into this

6(105/R) - 2R  = 12       multiply through  by R

630  -  2R^2  = 12R     rearrange

2R^2  + 12R  - 630 = 0      divide through  y  2

R^2  + 6R  - 315   = 0   factor

(R + 21) ( R - 15)  = 0 

Set  both factors to 0  and solve for R

R  = -21   reject

R  = 15   accept

So....there were 15  rows  in Formation A