CPhill

https://web2.0calc.com/questions/help-help_47

https://web2.0calc.com/questions/geometry_81168

https://web2.0calc.com/questions/counting-problem_70639

xy = 2/5

y = (2/5)x^(-1)

6x +  3  / (5 * (2/5) x^(-1))  =  

6x +  (3/2)x  

(15/2) x

There is no minimum......this is a line through the origin with a slope of  15 / 2

Construct a circle at B  with radius BC  that intercepts the semi-circle at E

Because BC = BE......then BE will be tangent to the semi-circle

Carry BE through to A ....let AD = x

OE = OC

BC = BE

BO = BO

Then triangles EOB and COB are congruent right triangles

tan angle OBC  = 1/sqrt 3  =  tan angle OBE 

tan ( OBC + OBE)  =  tan (ABC)....so.....

[ 2 * 1/sqrt 3 ] /  [1 - (1/sqrt 3)^2]  =   (2 + x) / sqrt 3

2/sqrt 3 * sqrt 3  =  ( 1 -1/3) (2 + x)

2  =  (2/3) (x + 2)

3 = x + 2

x =1

So

AB  = sqrt [ AC^2 + BC^2 ] = sqrt [ (2 + x)^2  + (sqrt3)^2 ]  = sqrt [ 3^2 + 3 ]  = sqrt 12  =  2sqrt 3

   PQ = 36, PR = 22, QR = 26 .....   This is all we need to know   

→  semi-perimeter  =  [ 36 + 22 + 26 ] / 2  = 42....using Heron's Formula

[ PQR]  =   sqrt [  42 *  6 * 16 * 20 ]     = 48 sqrt 35  ≈  284   

Take the derivative  with respect to x and  set it = 0

2x -16  = 0

x = 8

Take the derivative with respect to y and  set to 0

2y + 16  = 0

y = -8

The minimum is

(8)^2 + (-8)^2 - 4(8) + 2(-8) - 12(8) + 14(-8) + 26 =   -102

The slope of a perpendicular line  to y  = 3x + 1 through (7,1)  =  -1/3

The equation of this  line is   y = (-1/3)(x - 7) + 1  = (-1/3)x + 7/3 + 1   =  (-1/3)x + 10/3

To find the intersection point of these two lines

3x + 1 =   (-1/3)x + 10/3

3x + (1/3)x  = 10/3 -1

10/3 x = 7/3

10 x  = 7

x = 7/10 = .7

y = 3(7/10) + 1  =  31/10 = 3.1

The reflected point is   ( .7 - (7 - .7) , 3.1 - (1 - 3.1) )  = ( 1.4 (-5.6 , 5.2)

2. Lateral surface area  =   base perimeter * heighr  =

( 8 + 10 + 12 + 14 + 16) * (12)  =

60 * 12  =

720 mm^2

W      y    A       2        X

                                  x                            

                                  C

                                  x

Z      B        4             Y

angle ACX  + angle BCY =  90

Let

sin ACX = 2 / sqrt [2^2 + x^2]

cos ACX = x/ sqrt ;2^2 + x^2]

sin BCY = 4 /sqrt [4^2 + x^2]

cos BCY = x /sqrt [ 4^2 + x^2]

sin ( ACX + BCY) = sin 90

sin ACX cos BCY  + cos ACX sin BCY

2 / sqrt [ 2^2 + x^2[ * x / sqrt [ 4^2 + x^2]  +  x/ /sqrt [ 2^2 + x^2 ] *  4 /sqrt [ 4^2 + x^2]   =1

2x + 4x    =   sqrt [ 4 + x^2] sqrt [ 16 + x^2]           

6x  =  sqrt [ (4 + x^2) (16 + x^2)]                 square both sides

36x^2  = (4 + x^2) (16 + x^2)

36x^2 = x^4 + 20x^2 + 64

x^4 - 16x^2 + 64  = 0

(x^2 - 8)^2  = 0

x^2 - 8   = 0

x = sqrt (8)

2x = 2sqrt 8

Draw yB  perpendicular to ZY

yB  = 2sqrt 8

yA  = BY - AX  = 4 - 2  = 2

By the Pythagorean Theorem

yA^2  + yB^2  = AB^2

2^2  + (2sqrt 8)^2  = AB^2

4 + 32 = AB^2

36 = AB^2

AB =  sqrt 36  =  6