Answer:
Dimensions 5in x 10in x 10in will yeild a box with a max. volume of 500 cubic inches
Step-by-step explanation:
Volume = height x length x width
considering 'x' as the length of the square corners that has been cut out from the cardboard and also, that is height of the cardboard box.
square corners are cut out so that the sides can be folded up to make a box, cardboard sides would reduce by 2x
therefore,
V = x (30-2x) (30-2x) ---> eq(1)
V=( 30x - 2x²) (30-2x)
V= 900x- 60x² - 60x² + 4x³
V= 4x³ - 120 x²+ 900x
Taking derivative w.r.t 'x'
dV/dx = 12x² - 240 x +900
dV/dx = 4 (3x² - 60x +225)
For maximum dV/dx, make it equal zero
dV/dx = 0
so, 4 (3x² - 60x +225)=0
3x² - 60x+225=0 (taking 3 common)
x² - 20x + 75 =0
Solving this quadratic equation
x² - 15x -5x + 75 =0
x(x-15) - 5(x-15) =0
Either (x-15)=0
x=15
Or x-5=0
x = 5
if we substitute x=15 in eq(1), volume becomes zero.
therefore, x cannot be 15
When x= 5
eq(1)=>V = 5 (30-2(5)) (30-2(5))
V= 5 (10) (10)
V= 500 cubic inches,
Therefore, Dimensions 5in x 10in x 10in will yeild a box with a max. volume of 500 cubic inches
. We have our c value and our a value, so we will sub in those to find b. 
and 256 = 400 - b^2. -b^2 = -144, so b = 12. There you go!