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Friday, March 13, 2015

CITE BATCH 2015


CONGRATULATIONS  TO  ALL MY STUDENTS  CITE BATCH 2015   

1 Auman , Karen Coe
2 Ferolin , Pearl Aiko
3 Igcalinos , Kristine Mae
4 Gamuza , Chrystylle Joice
5 Misare , Ian
6 Acut , Harry vey
7 Galanto , Axl John
8 Cansa , Lalah
9 Agno , Angelica M.
10 Alsola , Robrich Neil
11 Amodia , Clarissa Mae
12 Aquino , Sheila Marie
13 Barlis , Christian
14 Benemerito , Jhon Ray M.
15 Benetiz , Angelu
16 Bucog , Johndel
17 Buenaventura , Kent Jude Q.
18 Buyco , Jessa Mae R.
19 Callar , Neil Christian V.
20 Canillas , Roan
21 Cantomayor , Queenee  A.
22 Carbonell , Leonard Jay F.
23 Cardines , Sheilacarm F.
24 Catipay , Leonardo Jr. M.
25 Daet , Felipe L.
26 Degamo , Mary Flor
27 Diesto , Ma. Aleli  G.
28 Ducot , Beverly Lyn
29 Flores , Judeleen D.
30 Garcia , Russell Kim
31 Garferio , Alvin Rey A.
32 Giango , Angel P.
33 Helaria , Jaycee John
34 Junas , Mary Grace
35 Kalis , Kevin Paul
36 Lao Bai Hanaria
37 Lavina , Franklin Jr.
38 Lesmoras , Fredelyn  B.
39 Liganad Jeric Von A.
40 Marquez , Bjorn Cyril
41 Matillano , Doris Joy D.
42 Palantang , Jayson
43 Palencia , Gilbeys T.
44 Panes , Aldrin B.
45 Patangan , Jaeckah C.
46 Pedrosa , Apple Jade L.
47 Pelayo , John S.
48 Pido , Japeth O.
49 Quemada , Jacques Genn C.
50 Quilaton , Eric U.
51 Reasonda , Michael
52 Rebuta Crisvegine  S.
53 Remorta , Reinalyn
54 Sacdalan , Belen
55 Sebuco , Ellen Rose
56 Sinarimbo , Myra Myca Apple
57 Sulilap Princess May
58 Tambis , Sherwin Clair
59 Vicente Mikko  C.
60 Yangzon , Jason G.
61 Reyes , Greycious Thea


Tuesday, March 10, 2015

TOP SCORERS IN F ELECT 3 FINALS


TOP SCORERS IN  F ELECT 3  FINALS

1.  AUMAN,  KAREN COE                                         98

2.  FEROLIN,  PEARL  AIKO                                     88
     IGCALINOS,  KRISTINE MAE                              88

3.  GAMUZA,  CHRISTYLLE                                      86

4.  MISARE, IAN                                                         82

5.  ACUT,  HARRY VEY                                             80

6.  GALANTO,  AXL JOHN                                         78

7.  CANSA,  LALAH                                                    74





Friday, February 27, 2015

ELLIPSE and HYPERBOLA

ELLIPSE

        An ellipse is a locus of a point which moves so that the sum of its distances from two fixed
     points is constant. The fixed points are called the foci of the ellipse and the line joining them
     is the principal axis.  

   An ellipse is symmetric with respect to its principal axis ( F’F ) and also with respect to the line B’B which is the perpendicular bisector of  F’F. The point of intersection O of B’B and F’F  is thus a point of symmetry and is called the center of the ellipse.

                                            PF’  +  PF  =  2a 

                                     Ö [x – (– c )]2  +  (y – 0)2     +     Ö (x – c)2 +  (y – 0)2          =    2a   

                                     Ö  (x + c )2  +  (y – 0)2       =    2a  –  Ö (x – c)2 +  (y – 0)2        

Square both sides of the equation

                             x2  +  2cx  +  c2   +  y2     =    4a2   –   4aÖ (x – c)2  +  y2      +   x2  –  2cx  +  c2  +  y2   

                          x2 – x2  +  y2 – y2 + c2 – c2 + 2cx +  2cx  –  4a2   =    – 4aÖ (x – c) +  y2       

                                                           4cx  –  4a2   =    – 4aÖ (x – c) +  y2        

                                              Divide all terms by 4, then square both sides again

                                     cx  –  a2   =    – aÖ (x – c) +  y2       

                     c2x2  – 2a2cx  +   a4   =  a2 (x2 – 2cx  +  c +  y2)    
 
                     c2x2  – 2a2cx  +   a4   =   a2x2  –  2a2cx   + a2c2 +  a2y2      

                    c2x2  – a2x2   –   a2y2   =  – a4  + a2c2  – 2a2cx  +  2a2cx

                              – x2(a2 – c2)  –  a2y2   =  – a2( a2 –  c2 )



                                                   but by Pythagorean theorem   a2 –  c2  =  b2
                                                         
                                           – x2 b2  –  a2y2   =  – a2b2 

                                                            Divide all terms by   – a2b2 

                                                x2      +     y2      =   1
                                               a2              b2            
                      
                       This is the standard equation of an ellipse with center at the origin and

                             ( x – h )2   +   ( y – k )2     =  1         for ellipse with center at the point  (h, k). 
                                 a2                    b2

           where,  a  is the length of the semimajor axis and  b  is the length of the semiminor axis.

              If  a  >  b , the axis of the ellipse is horizontal and  if  a < b ,  the axis is vertical.
 

                c  =  √ a2 – b2   =  the distance from the center of the ellipse to the foci = CF = CF’

                b  = the distance from the center of the ellipse to the minor vertices  = CB = CB’

                a  =  the distance from the center of the ellipse to the major vertices  = CV = CV’


          The equation  Ax2 + Bxy + Cy2 + Dx + Ey + F = 0 is a general equation of an ellipse if 
           B = 0  and A  ¹  C and  both A and C have the same sign.


                Example
                     Find the center, vertices and foci of the given ellipse.

                       1.   x2  +  4y2  =  16                                 a > b , axis is horizontal

                             x2  +  4y2  = 16                                  a =  4 ,   b = 2  and      c = √ 12    
         
                             x2   +    y2   =  1   .                             center, C( 0 , 0 )      
                            16          4                                           V( 4, 0 )  and  V’ (– 4 , 0 )

                       ( x – 0 ) 2   +   ( y – 0 )2   =  1   .              B ( 0 , 2 )  and  B’( 0 , – 2)
                                 42                    22                              F (√ 12 ,  0 )   and  F’ (– √ 12 ,   0 )

                                                      Graph of the ellipse   x2  +  4y2  =  16  

                                              

                   Do these exercise.  Find the center, foci, major and minor vertices of the following ellipse
                    and draw the graph.

                                   1.    x2  +  4y2 – 8y  =  12                              

                                   2.    x2  +  4y2 – 6x  =  7                                

                                   3.    x2  +  4y2 – 4x – 8y  =  0           
             
                                   4.    9x2  +  4y2  =  36

                                   5.   16x2  + 25y2 – 128x – 150y + 381 = 0

                                   6.   25x2  + 16y2 – 100x – 32y  =  284



                     HYPERBOLA
           A hyperbola is  locus of point which moves so that the difference of its distances from two fixed
           points is constant. The fixed points are called the foci of the hyperbola and the line joining them is
          the principal axis. The segment V’V is the transverse axis having length of 2a and the segment B’B
         is the conjugate axis whose length is 2b. The segment F’F is the principal axis and is equal to 2c.
           
                                                                   c2  =  a2  +  b2 

                      Equation of Hyperbola      

                            The general equation of  hyperbola is defined by

                                                    Ax2 + Cy2 + Dx + Ey + F = 0

                                                    where  A and C have unlike signs.  


                            The standard equation of a hyperbola is given by

                                      a) with C ( 0 , 0 )

                         x 2    –     y 2    =  1           For hyperbola with horizontal transverse axis.
                        a2            b2

                          y 2    –     x 2    =  1           For hyperbola with vertical transverse axis.
                     a2            b2

                        b) with C ( h , k )

                          ( x – h ) 2   –   ( y – k ) 2   =  1     For hyperbola with horizontal transverse axis.
                         a2                    b2

                      ( y – k ) 2   –   ( x – h ) 2   =  1      For hyperbola with vertical transverse axis.
                               a2                   b2


                          Example 1.    x2  –  4y2  =  16                                           
              

                              Solution :
                          
                                          x2  –  4y2  = 16                                               
         
                                            x2    –    4y2     =    16                                      
                                           16           16           16

                                             x2    –     y2    =  1                                                    
                                            16           4          

                                     ( x – 0 )2    –   ( y – 0 )2     =  1                           
                                           42                    22                              
                                     a =  4 ,   b = 2  and    c = √ 16  +  4     =  √20 


                                     C ( 0 , 0 )  ,   axis is horizontal


                                    V ( 4 , 0 ) ,  V’( – 4 , 0 ) ,    
                      
                                     F (√20 , 0 ) ,  F’(–√20 , 0 ) , 

                                    B( 0 , 2 ) ,  B’( 0 , – 2)     


              Do these exercises. 
                Find the center, conjugate and transverse vertices,  foci and asymptote of the given hyperbola
                and draw the graph.

                               1.    x2  –  4y2 – 8y  =  12                             

                               2.    y2  –  4x2  +  6x  =  7                             

                               3.  x2  –  4y2 – 4x – 8y  =  0                         

                              4.   4x2  –  9y2  =  36

                              5.  16y2  – 25y2 – 128x – 150y + 381 = 0

                              6.   16y2  –   25x2  – 100x – 32y  =  284

                              7.   9x2  –  4y2   =  36

                              8.   9x2  –  18x  –  4y2  +  8y  =  31
 
                              9.   4x2  –  9y2  + 18y   =  45

                            10.   9x2  +  18x  –  4y2   =  27 



Thursday, February 5, 2015

CIRCLES and PARABOLA





CIRCLE

A circle is a locus of a point which moves in a plane so that its distance from a fixed point remains constant. The fixed point is called the center of the circle and the constant distance is the radius of the circle.

            By the definition of a circle, we have  CP  =  r,  or
              
                        Ö ( x – h)2 + ( y – k )2           =  r

           By squaring both sides of the equation, we have

                                ( x – h )2 + ( y – k )2  =  r2  

      which is called the standard form of the equation of a circle.

         Expanding the equation we have,

                             x2 – 2hx + h2 + y2 – 2ky + k2  =  r2,  or

                             x2  +  y2  – 2hx  - 2ky  + ( h2 + k2 – r2 )  =  0. 
                        If  -2h = D ,  -2k = E ,  (  h2 + k2 – r2 ) = F       

        Then this equation is of the form    

                             X2 + y2 + Dx  +  Ey  +  F  =  0 ,

         which is the general  form of the equation of a circle.
Given a general equation of a circle, we need to transform it into the standard form so the
center and the radius can be determined and the graph can be drawn.
                                    
Example.
   1.  Find the center and the radius of the circle  x2 + y2  =  16.
                  Solution :
                                    x2 + y2  =  16

                                    ( x – 0 )2 + ( y – 0 )2 = 42 

                                    C ( 0, 0 ),   r = 4 

  2.  Find the center and the radius of the circle 
         x2 + y2 + 4x – 2y  =  4  and draw the graph.
                  Solution :
                        x2 + y2  +  4x – 2y  =  4

                       x2 + 4x  + 4  + y2 – 2y + 1  = 4 + 4 + 1

                      ( x + 2 )2  +  ( y  – 1 )2  =  32   

                        C ( – 2,  1 ) ,   r  =  3

Exercises :
Find the center and the radius of the following circle and draw the graph.
 1.  x2+y2 + 4x – 6y–12  = 0          6.  x2+ y2 – 10x + 9 = 0     

 2.  x2 + y2 – 4x = 5                       7.  x2 + y2+ 10x –  6y = 2 

 3.  x2 + y2 + 10y – 11 = 0             8.  x2 + y2 – 16y +15 = 0   

 4.  x2 + y2 –10x + 6y = 2              9.  x2+ y2 +10x – 4y –7 = 0  

 5.   x2 + y2 +14y + 24 = 0           10.  x2+ y2 – 4x + 6y –12 = 0


    

CIRCLES DETERMINED BY THREE CONDITIONS

Find the equation of the circle that passes through the points (-1,2),
(1,1) and (2,3).
Two solutions are available for the problem
   a) First solution is by using the general equation.
   b) Second solution is by using standard equation.
       To find the center of the circle use the intersection of the  
       perpendicular  bisector, then use distance formula to find the  
       radius of the circle.

 Solution by the general equation
       at (-1,2) :   (-1)2 + 22 – D  + 2E  + F  = 0
                                       5 – D + 2E + F = 0    ¬ EQ 1

      at (1,1) :    12  +  12  +  D  +  E  +  F  =  0    
                                    2  +  D  +  E  +  F  =  0  ¬ EQ 2

      at (2,3) :  22  +  32  + 2D + 3E + F = 0
                                 13 + 2D + 3E + F = 0   ¬ EQ 3 

                      EQ 2 :  2 + D  +  E +  F  =  0
                  +  EQ 1 :  5 – D + 2E + F = 0                 .     
                                   7    + 3E  + 2F = 0   ¬ EQ 4  

                    2(EQ 1 ) :  10 – 2D + 4E + 2F  = 0
                     +  EQ 3 :   13 + 2D + 3E + F = 0     .        
                                       23  + 7E  + 3F  = 0  ¬ EQ 5   

                           3( EQ 4 ) :    21   +  9E  + 6F  = 0 
                       – 2( EQ 5 ) :  – 46 – 14 E – 6F = 0   .       
                                              – 25 – 5E  =  0
                                                           E = – 5 
                            EQ 4 :   F = ½ [ – 7 – 3(– 5)]  =  4   
                            EQ 2 :  D = – 2 – E – F 
                                        D = –2 – (–5) – 4  =  – 1
        Therefore, the equation is  x2 + y2 – x – 5y + 4 = 0.

   b) Solution by using the standard equation

                      Perpendicular bisectors :
                     ( x – 1 )2 + ( y – 1 )2  = ( x – 2 )2  + ( y – 3 )2
             x2 – 2x + 1 + y2 – 2y + 1  =  x2 – 4x + 4 + y2 – 6y + 9
                        – 2x + 4x – 2y + 6y = 13 – 2
                                  2x + 4y = 11  ¬ EQ 1

                       ( x + 1 )2 + ( y – 2)2  = ( x – 1 )2 + ( y – 1 )2  
               x2 + 2x + 1  + y2 – 4y + 4 = x2 – 2x + 1 + y2 – 2y + 1  
                            2x + 2x – 4y + 2y = 2 – 5
                              4x – 2y  = – 3   ¬ EQ 2

                           EQ 1 :  2x + 4y = 11  
                     2( EQ 2) :   8x – 4y = – 6        .    
                                       10x  =  5
                                            X = ½

             EQ 2 :  y = ½ (4x + 3 ) = ½ [ 4( ½ ) + 3 ]  =  5/2  

                                        C( ½ , 5/2 )

                        r2  =  ( 1 – ½ )2  + ( 1 – 5/2 )2  =   5/2

          By the standard equation of a circle :

                      ( x – ½ )2  +  ( y – 5/2 )2  =  5/2

                        x2 – x  + ¼  + y2 – 5y  + 25/4  =  5/2

                        x2 – x + y2 – 5y  + ¼ + 25/4 – 5/2  = 0

                        x2  –  y2  –  x  –  5y  +  26/4 –  10/4  = 0   

                        x2  –  y2  –  x  –  5y  +  4  =  0  
                    
   Therefore the equation of the circle is  x2 – y2 – x – 5y + 4  = 0


   Application problems.
     Find the equation of the circle
      1.  with r = 5 and C( 2 , -5 ).
      2.  containing ( -4 , 3) and C( 4 , -2 )
      3.  tangent to the x-axis, C( 3 , 2 )
      4.  having (-1, -2) and (3, 4) as end points of the diameter.
      5.  tangent to the y-axis with C( -4, 3 ).
      6.  passing through ( 2,3 ), (6,1) and (4,-3)
      7.  passing through ( -3 , 1 ),  ( 5 , -3) and ( -2 , 4 ).
      8.  center on the y-axis and passes through the origin and (4, 2).
      9.  tangent to the axes and  C( -5, -5).
    10.  tangent to the axes and  r = 4.   ( 4 answers )
    11.  r = 5, tangent to the line 3x + 4y = 24 at ( 2 , 9/2)
    12.  C( 3, 5) and containing  the origin.
    13.  Find the equation of the circle that passes through the points
            ( -1 , 2 ),  ( 1 , 1 )  and ( 3 , 2 ).
    14.  Find the equation of the circle that passes through the point
            ( 9 , 7 ) and is tangent to both the y-axis and the
            line  3x – 4y  =  24.
     15.  Tangent to the axes,  center 4th quadrant,  radius 3.




RADICAL  AXIS 
     If  x2 + y2 + Dx + Ey  + F = 0   and  x2 + y2 + D’x + E’y + F’ = 0,
are the equations of two non-concentric circles, then the equation
  x2 + y2 +  Dx + Ey + F + k(x2 + y2 + D’x + E’y + F’ ) = 0,   ( EQ 1 )   
 represents a circle for all values of k except  k = –1. If the given circles intersect 
in two distinct points, all members of the family 
( EQ 1 ) will pass through these points. This is evident since any point (x1, y1
that satisfies the given equations will make the left side of (EQ1) read 0 + k0, 
which is zero for any value of k. If the given circles are tangent at a point,  the 
family ( E Q 1)  will be tangent  to them  at  the point  of tangency.

 Example.
   Find the equation of the circle that passes through the points of intersection of the circle  x2 + y2 = 2x,  x2 + y2 = 2y,  and contains the point  ( a )  ( 2, 3 )   and  ( b )  ( 2, 1 ).  

          By  EQ 1 :     x2 + y2 – 2x  +  k( x2 + y2 – 2y ) =  0

          at (2 , 3) :     22  + 32 – 2(2) + k [ 22 + 32 – 2(3)] = 0

                                 4 + 9 – 4 + k [ 4 + 9 – 6 ] = 0

                                              9 + k ( 7 ) = 0

                                                    k =  – 9/7

       The member of this family that passes through ( 2, 3 ) is

                          x2 + y2 – 2x  – (9/7)( x2 + y2 – 2y ) =  0

                        7x2 + 7y2 – 14x  – 9x2 – 9y2 +  18y =  0

                          – 2x2 – 2y2  – 14x + 18y = 0

                               x2  + y2  +  7x – 9y  =  0


     
  In  EQ 1,  if we set  k = – 1, the linear equation

                    ( D – D’) x  +  ( E – E’ )y + ( F – F’) = 0  is obtained.

     This equation is called the radical axis of the two circles and has the following properties :
      1.  If two circles intersect in two distinct  points, their radical axis  
           contains the common chord of the circles.
      2.  If two circles are tangent, their radical axis is the common tangent to the circles at their point of tangency.
      3.  The radical axis of two circles is perpendicular to their line of
           centers.
      4.  All tangents drawn to two circles from a point on their radical axis have the same length.



  Find the radical axis of the following circles.

    1.  x2  + y2  +  4x – 8y  = 5  and  x2  + y2  +  10x – 6y  = 2
    2.  x2  + y2  – 4x + 8y  = 16  and  x2  + y2  – 10x + 6y + 9  =  0
    3.  x2  + y2  +  4x  = 12  and  x2  + y2 – 6y  = 7
    4.  x2  + y2  – 8y  = 9  and  x2  + y2  +  10x  =  0
    5.  x2  + y2 + 8y  =  9  and  x2  + y2  +  10x – 6y  =  2




PARABOLA

Parabola is the locus of a point which moves so that its distance from a fixed point and a fixed line are equal.

The fixed point is called the focus of the parabola and the  line is called the directrix of the parabola.


      DD’  –  directrix
      E, E’  –  end points of the latus rectum
      V  –  vertex of the parabola
       F – focus of the parabola 
      VF = a   –  focal distance
       DD’V =  VF =  a 
      EF = FE’  =  2a
  The axis of the parabola is horizontal ( OX ).

From the above figure and by definition of parabola :

 PF  =  PQ  

Ö ( x – a )2 + ( y – 0 )2    =  ( x + a ) 

Square both sides

( x – a )2 + y2  = ( x + a )2

X2 – 2ax + a2  + y2  =  x2  + 2ax  + a2

X2 – x2 + a2 – a2 + y2 = 2ax + 2ax

y2 =  4ax 
 
Standard equation of a parabola having horizontal axis and opens to the right.        
                                                          
From the figure above and by definition of parabola :

PF  =  PQ

Ö( x – 0)2 + ( y – a )2   =  y + a

Square both sides

X2 + y2 – 2ay + a2  =    y2 + 2ay + a2

X2 + y2 – y2 + a2 – a2  =  2ay  + 2ay

X2 = 4ay

Standard equation of a parabola with vertical axis (OY) and opens upward.

General Equation of a Parabola

1.  x2 + Dx + Ey + F = 0 ,  E ¹ 0,   with vertical axis ( OY )

2.  y2 + Dx + Ey + F = 0,   D ¹ 0,  with horizontal axis (Ox )

Standard Equation of the Parabola

A. with vertex at V( 0, 0 )

      1.  y2 =  4ax  ----   opens to the right

      2.  y2 =  – 4ax  ----  opens to the left

      3.   x2 = 4ay  ----  opens upward

      4.   x2 =  – 4ay  ----  opens downward  
     
B. with vertex at  V( h, k )

      1.  ( y – k )2 =  4a( x – h )    ----   opens to the right

      2.  ( y – k )2 =  – 4a( x – h)  ----  opens to the left

      3.   ( x – h )2 = 4a( y – k )   ----  opens upward

      4.  ( x – h )2 =  – 4a( y – k   ----  opens downward  

Examples.

Draw the graph of the following parabola.
  1.  2x2 – 8y = 0                 3.  x2 + 2x – 6y = 2
  2.  3y2 + 18x = 0               4.  y2 + 2y – 6x = 2 
 5.  x2 + 4y – 4 = 0                      8.  y2 + 8x  + 16 = 0
 6.  y2 – 4x + 4 = 0                      9.  x2 = 4x + 4y 
 7.  y2 + 2x + 6y + 17 = 0          10.  x2 – 2x + 2y + 7 = 0



QUADRATIC FUNCTIONS

        A function of the form y = ax2 + bx + c,  where  a ¹ 0 is a quadratic function. Since the equation has the form of the general equation of a parabola  x2 + Dx + Ey + F = 0, the graph of every quadratic function is a parabola with a vertical axis. It is concave upward if a is positive or concave downward if a is negative.
         At the vertex of the parabola,  x = – b/2a. This value of x gives the function its minimum value if a is positive or concave downward if a is negative. The equation of the axis of the parabola is x = – b/2a.   
        As x increases, many functions increase to a maximum value, decreasing thereafter, or decrease to a minimum and begin to increase. In such cases, it is a problem of prime importance to determine the maximum or minimum value. To solve problems for functions in general, differential calculus is required. For the quadratic function it may evidently be solve by merely finding the vertex of the parabola, since the ordinate of the vertex is the greatest or least ( algebraically ) of all the ordinates, according as the parabola opens downward or upward.
STEPS IN GRAPHING A QUADRATIC FUNCTION
  1. Compute x = – b/2a, and then the corresponding value  of y
      to obtain the coordinates of the vertex of the parabola.
       
  2. If the value of  x = – b/2a is convenient, form a table by using
      pairs of values of x where each pair, the values are
      equidistant from the vertex on either side. The value of y
      corresponding to any pair will be equal.

  3. If the value of  x = – b/2a  is inconvenient, form the table by
      choosing values of x arbitrarily on either side of  x = – b/2a.

Example.  Determine the critical points whether it is maximum or minimum.
                1.  y  =  x2 – 10x + 25 
                2.  y =  4x –  x2                                               
                3.  y =  2x2 – 10x + 3
                4.  y = 15x – 3x2  
                                                          
Application problems.

Use quadratic function to evaluate the following problems. 
 1. Find the dimensions of the largest rectangular lot that can be
     enclosed by 100 m of fencing material.
      
 2. A rectangular field is to be fenced off along the bank of a
   river. If no fence is needed along the river, what is the shape
  of the field requiring 400 m of fence if the area is to be
  maximum?

   1. Solution
       Let  x and y be the dimensions (length and width) of the lot       
            Area         :   A = xy
               Perimeter :    2x + 2y = 100                                      
                                x +  y = 50                            
                                y = 50 – x 
                            x(50 – x ) = A                                                                                             
                             50x – x2  =  A                                 

                           Standardize the equation :
                               X2 – 50x = – A                                                                
                            x =  – b/2a  = – (–50) / 2 ( 1 )
                            x =  25 m  
                            y = 50 – x
                            y = 50 – 25
                            y = 25 m

                  Alternately :       

                                By completing the square :
                              x2 – 50x + 625 = – A + 625  
                             ( x – 25 )2 = – ( A – 625 )
                             x – 25 = 0        |   A – 625 = 0
                             x = 25 m          |   A = 625 m2   
                            y = 50 – 25
                              = 25 m

               The dimensions are 25 m by 25 m.  
               The maximum area is  625 m2       

           
  B.  Use quadratic functions to solve the problems. 
     1. A rectangular field is to be enclosed, and divided into three
        lots by fences parallel to one of the sides. Determine the
      dimensions of the largest field that can be enclosed by
      2,400 m of fencing material.
  
    2.  A rectangular lot  is to be fenced off along a highway. If the
         fence along the highway cost  P 150 per meter, and on the
         other sides  P 100 per meter, determine the dimensions of
         the largest lot that can be enclosed  for a budget of
         P 100,000.
  
   3. A rectangular field is to be fenced off along the bank of a
       river. If no fence is needed along the river, what is the shape
      of the  field requiring 400 m of fence if the area is to be
      maximum?