These two max min problems involve shapes with surface area, volume, etc.
The first one is asking for a ratio, not just a simple value. The second one says quadrilateral but does that mean it is a cube or something?
let x and y be the lengthes of the diagonals. Then the area of the quadrilateral is calculated by:
a = ½ * x * y
From the given conditions you know: x + y = 6 ==> y = 6 - x
Plug in the term for y into the first equation:
a(x) = ½*x*(6 - x) = -½x² + 3x
The graph of this function is a parabola opening downwards. The maximum value is at it's vertex: V(3, 4.5)
You get: x = 3, y = 3 the maximum area is 4.5 cm²
The quadrilateral is a square with diagonals which are 3 cm long.
volume of the can: V = k = π*r²*h ==> h = k/(π*r²) 
complete surface area: a = 2*πr² + 2*π*r*h
costs: c = 2*πr²*0.03 + 2*π*r*h*0.06  Plug in the term for h from  into :
c(r) = 2*πr²*0.03 + 2*π*r*k/(π*r²)*0.06 = 0.06*πr² + 0.12k/r
You'll get an extremum if the first derivation equals zero:
c'(r) = 0.12πr - 0.12k/r² With c'(r) = 0 you get r = ³√(k)
(It is necessary to prove if you've got a minimum. I leave this to you )
Now plug in this value into  and calculate h/r = 1/π