24
32
45
60
Let the sides of the rectangle be 15k and 8k, and the side of the square be x. Then (15k - 2x)(8k - 2x)x is the volume.
→v = 2(2x³ - 3kx² + 60k²x)
For maximum volume, the derivative of V w.r.t. x should be zero.
→dv/dx|x=5 = 0
6x² - 6kx + 60k²|x=5 = 0
6k² - 6k + 15 = 0
k = 3, k = 5/6
Only k = 3 is permissible. So, the sides are 45 and 24.