let

l = the length

w = the width

h = the height

the volume of the box V = length*width*height

l*w*h = 570

the area of the surface A = 2wl + 2hl + 2lw

2wl + 2hl + 2lw = 478 divide both sides by 2

wl + hl + lw = 239

we have 2 equation and 3 variables follows there are infinitely many possible solutions.

We will find one possible solution using trials:

since 570 = 2*3*5*19 lets assume that h = 5 then we have

5lw = 570 divide both side by 5

lw = 570/5

lw = 114

and also for wl + hl + lw = 239 and h = 5 we get

5(l + w) + lw = 239

plug lw = 114 into the last equation

5(l + w) + 114 = 239

l + w = (239 - 114)/5

l + w = 25

l = 25 - w

if we plug l = 25 - w into lw = 114 we get

(25 - w)w = 114

by solving we find

w1 = 19

w2 = 6

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for w1 = 19

l1 = 25 - 19 = 6

for w2 = 6

l2 = 15 - 6 = 19

since length >= width we found the following dimensions of the box

height = 5

width = 6

length = 19

indeed V = 5*6*19 = 520 and A = 2(5*6 + 5*19 + 6*19) = 478.