Answer :
The INSIDE dimensions of the rectangular safe :
Length L = 1.20 m Breadth B = 1.20 m Height H = 2.00 m
Let thickness of safe be t meters
To get the outside dimensions: add t on each side of a dimension.( left side and right side for widht and length. Also at top and bottom for height). so each dimension becomes plus 2t.
Outside Length = 1.20+2t width = 1.20+2t height = 2.00+2t
Ouside volume = (1.2+2t)² (2+2t) = 2²(0.6+t)² 2 (1+t)
The volume of Steel used is the difference of the outside volume and inside volume.
Hence : 8(t²+1.2t+0.6²)(1+t) - 1.2² * 2 = 1.25 meter³
8 (t² + 1.2 t + 0.36 + t³ + 1.2t² + 0.36 t) - 2.88 = 1.25
8 t³ + 17.6 t² + 12.48 t + 2.88 - 2.88 = 1.25
find t by solving the equation. The roots of this polynomial could be possibly
factors of +- 1.25 / factors of 8
t = 0.0886 meters or 0.0887 meter , that is 8.86 cm or 0.0887 cm approx.
t is so small that t³ is very small compared to t² and t. SO neglect the t³ term and solve the equation using te quadratic equation formula. The answer u get will be very close the actual one.
Length L = 1.20 m Breadth B = 1.20 m Height H = 2.00 m
Let thickness of safe be t meters
To get the outside dimensions: add t on each side of a dimension.( left side and right side for widht and length. Also at top and bottom for height). so each dimension becomes plus 2t.
Outside Length = 1.20+2t width = 1.20+2t height = 2.00+2t
Ouside volume = (1.2+2t)² (2+2t) = 2²(0.6+t)² 2 (1+t)
The volume of Steel used is the difference of the outside volume and inside volume.
Hence : 8(t²+1.2t+0.6²)(1+t) - 1.2² * 2 = 1.25 meter³
8 (t² + 1.2 t + 0.36 + t³ + 1.2t² + 0.36 t) - 2.88 = 1.25
8 t³ + 17.6 t² + 12.48 t + 2.88 - 2.88 = 1.25
find t by solving the equation. The roots of this polynomial could be possibly
factors of +- 1.25 / factors of 8
t = 0.0886 meters or 0.0887 meter , that is 8.86 cm or 0.0887 cm approx.
t is so small that t³ is very small compared to t² and t. SO neglect the t³ term and solve the equation using te quadratic equation formula. The answer u get will be very close the actual one.