CPhill

(5/3)x   + 12/7   = y

5/3  is a  terminating decimal   while   12/7  is  not......so   it will not pass through any lattice points 

I also believe that there is  no minimum distance to any lattice point.....

The minimum distances will approach zero but they never converge to any particular value 

Here's what I think :

This will be a right triangle with legs = 2  and hypotenuse = 2sqrt (2)

Minimum perimeter =   2 + 2 + 2sqrt (2) =   4 + 2sqt (2)  

The region is  unbounded

See here :  https://www.desmos.com/calculator/m5ajeuqdn4

f(x) = 2x^2 + 22x - 76

2x^2  + 22x - 76    =     a

2x^2 + 22x  - 76 - a  =  0

2x^2 + 22x - (76 + a) = 0

Take the discriminant and set to 0

(-22)^2  - 4 (2) [ -76 - a ] = 0

484 + 608 + 8a  =0

1092  = -8a

a = 1092/ -8  =   -136.5

(The value of x  =  5)

(In other words, f(5)  = -136.5  )

If arc LM  = 90    then angle  LOM  =90  because it is a  central agle in the larger circle

Then LOM is a right triangle with legs OM , LM  =1

[ LOM ]   =  (1/2) (OM) (LM)   =   (1/2) (1)(1)   = 1/2

Area of large square   =  (1 + 1.5)^2  = (2.5)^2  = 6.25

2k - 3(-3)^2  + 7  =  -10  +  4 ( 2k -3(-5)^2  + 7)

2k - 27 + 7  = -10   + 4 ( 2k - 75 + 7)

2k -27 + 7 =  -10 + 8k - 300 + 28

2k -20  =  8k - 282

262  = 6k

262 / 6  =  k  =  131 / 3

radius =  (9 + 16 + 9) /  2  =   17

Let O  be the center of the circle

QO =  8  OD  = 17

Triangle QOD is right  with OD=  the hypotenuse

So

DQ = sqrt ( 17^2 - 8^2) =  sqrt ( 225) =  15

So ABCD is a trapezoid  with height  =  15    and bases of 34  and 16

[ ABCD ] =  (1/2) ( 15) ( 34 + 16)  =   375

4 feet  =  48  inches = width

Each can has a diameter=  2 in

So

48/2 =   24 cans can  be  packed in one row

And 23  can be packed in the next row

15 in / 5 =    3  =  number of  cans in a  stack

Each two rows   of cans occupy   [10 + 5sqrt (3)] in  of the length

So     integer ( 24  / [ 10 + 5sqrt 3 ]  )  =     1  two row unit

This means only  two rows of cans can  be fitted  lengthwise   

Total number of cans  =  3 ( 24 + 23)  =  147

Surface area =

pi  [ L (R + r)   ]  +   pi ( R^2 + r^2 )             where  L = sqrt [H^2 + (R - r)^2 ]

So

pi [ sqrt ( 4^2 + (12 - 9)^2 ) ( 12 + 9 )]  +  pi (12^2 + 9^2) ] =

pi [ sqrt ( 16 + 9)  *  (21)  ]  +  pi ( 225) =

pi [ 5 * 21 + 225 ] =

330 pi

Volume  =

pi  (H /  3 )   *   ( R^2  + Rr + r^2)

So

pi (4/3) * (12^2 + 12*9 + 9^2)   =

pi (4/3) * ( 144 + 108 + 81)  =

pi (4/3) ( 333) = 

444 pi

A + V =    (330 + 444) pi  =   (774) pi