Answer:
Width of box= 2inches
Length of box= 4inches
Height of box= 6inches.
Step-by-step explanation:
Let width of box=x inches
Length of box = twice of width=[tex]2\times x[/tex]=[tex]2x[/tex]
Height of box= 4 inches greater than width= [tex]x+4[/tex]
Volume of box= 48 cubic inches
We know that the formula of volume of cuboid= [tex] length\times breadth\times height[/tex]
Apply the formula
Volume of box= [tex]x\times 2x\times (x+4)[/tex]
Volume of cube = [tex]2x^2(x+4)[/tex]
[tex]2x^2(x+4)=48[/tex]
[tex]x^2(x+4)=24[/tex]
[tex]x^3+x^2-24[/tex]
Apply inspection method to solve the equation
Put [tex]x=0[/tex]
Then we get [tex]-24\neq0 [/tex]
Hence, x=0 is not the solution of x
Put x=1 in the equation then we get
[tex]-22\neq 0[/tex]
Hence x=1 is not the solution of equation.
Put x=2 then we get
[tex](2)^3+(4)^2-24[/tex]
8+16-24=0
Hence, x=2 is the solution of equation .
[tex] (x-2)(x^2+6x+12)[/tex]=0
Now substitute equation [tex]x^2+6x+12[/tex]=0
Sum roots =6
Product of roots=12
When sum of roots is greater than zero and product of roots is greater than zero then value of roots of equation is imaginary.
Hence, the roots of equation [tex]x^2+6x+12=0[/tex] are imaginary.
Lenght , widht and height are dimensions of box therefore, imaginary value are not possible.
Hence,[tex] x=2 [/tex] is the only real values of root of equation .Therefore, it is possible and other two imaginary value of roots are not possible .
Widht of box=2 inches
Length of box = [tex]2\times2[/tex]=4inches
Height of box=[tex]x+4[/tex]=2+4=6 inches
