Answer :
This question I answered earlier. I will like to solve in an easier method which u can follow in such problems with cubic equations. It is by successive iteration.
The INSIDE dimensions of the rectangular safe:
Length L = 1.20 m Breadth B = 1.20 m Height H = 2.00 m
Inside volume = 1.2² * 2 = 2.88 m³
Let thickness of safe be t meters
To get the outside (external) dimensions: add t on each side of a dimension (left side and right side for width and length. and at top and bottom for height). So each dimension becomes plus 2t.
Outside Length = 1.20+2t External width = 1.20+2t external height = 2.00+2t
External (Outside) volume = (1.2+2t)² (2+2t) = 2²(0.6+t)² 2 (1+t)
The volume of Steel used is the difference between the outside volume and the inside volume.
Hence : 8 (t+0.6)² (t+1) - 2.88 = 1.25 meter³
(t+0.6)² (t+1) = 4.13 / 8 = 0.51625
Let (t + 0.6) = x
x² (x+0.4) = 0.51625 or x³ + 0.4 x² - 0.51625 = 0
We can try the ways of factorization and finding zeros. But here I show another way of successive iterations to solve. rewrite the equation as,
x² = 0.51625 / (x+0.4) or x = √[0.51625 / (x+0.4)]
x = 0.7185 / √(x+0.4)
We will find the value of x by starting with some value and calculating it many times, so that we get the precision we want. Substitute a value of x on R H S and find the L H S
Start with x = 1 , new value of x = 0.7185 / √(1+0.4) = 0.6072
use x = 0.6072, new value of x = 0.7185 / √(0.6072+0.4) = 0.7159
next value of x in iteration = 0.7185 / √(0.7159+0.4) =0.68016
Next iteration x = 0.6913
next iteration x = 0.68778
next iteration x = 0.6889
next iteration x = 0.68854
next iteration x = 0.68866
next iteration x = 0.6886...
Let us take x = 0.6886 as these digits are repeated in the next iteration.
t = x - 0.6 = 0.0886 meters
or t = 8.86 cm approximately
Verification of answer :
Outside volume = (1.2+2*0.0886)² (2+2*0.0886) = 4.1294 m³
Inside volume = 1.2² * 2 = 2.88 m³
Volume of steel = 1.2494 m³ ≈ 1.25 m³
================================
as x³ + 0.4 x² - 0.51625 = 0
x ³ = (0.51625 - 0.4 x²)
x = ∛(0.51625 - 0.4 x²)
we can use this equation also for successive iterations.
The INSIDE dimensions of the rectangular safe:
Length L = 1.20 m Breadth B = 1.20 m Height H = 2.00 m
Inside volume = 1.2² * 2 = 2.88 m³
Let thickness of safe be t meters
To get the outside (external) dimensions: add t on each side of a dimension (left side and right side for width and length. and at top and bottom for height). So each dimension becomes plus 2t.
Outside Length = 1.20+2t External width = 1.20+2t external height = 2.00+2t
External (Outside) volume = (1.2+2t)² (2+2t) = 2²(0.6+t)² 2 (1+t)
The volume of Steel used is the difference between the outside volume and the inside volume.
Hence : 8 (t+0.6)² (t+1) - 2.88 = 1.25 meter³
(t+0.6)² (t+1) = 4.13 / 8 = 0.51625
Let (t + 0.6) = x
x² (x+0.4) = 0.51625 or x³ + 0.4 x² - 0.51625 = 0
We can try the ways of factorization and finding zeros. But here I show another way of successive iterations to solve. rewrite the equation as,
x² = 0.51625 / (x+0.4) or x = √[0.51625 / (x+0.4)]
x = 0.7185 / √(x+0.4)
We will find the value of x by starting with some value and calculating it many times, so that we get the precision we want. Substitute a value of x on R H S and find the L H S
Start with x = 1 , new value of x = 0.7185 / √(1+0.4) = 0.6072
use x = 0.6072, new value of x = 0.7185 / √(0.6072+0.4) = 0.7159
next value of x in iteration = 0.7185 / √(0.7159+0.4) =0.68016
Next iteration x = 0.6913
next iteration x = 0.68778
next iteration x = 0.6889
next iteration x = 0.68854
next iteration x = 0.68866
next iteration x = 0.6886...
Let us take x = 0.6886 as these digits are repeated in the next iteration.
t = x - 0.6 = 0.0886 meters
or t = 8.86 cm approximately
Verification of answer :
Outside volume = (1.2+2*0.0886)² (2+2*0.0886) = 4.1294 m³
Inside volume = 1.2² * 2 = 2.88 m³
Volume of steel = 1.2494 m³ ≈ 1.25 m³
================================
as x³ + 0.4 x² - 0.51625 = 0
x ³ = (0.51625 - 0.4 x²)
x = ∛(0.51625 - 0.4 x²)
we can use this equation also for successive iterations.