Question

Find the dimensions of the box described. The length is twice as long as the width. The height is 4 inches greater than the width. The volume is 48 cubic inches. Find the length, width, height

Answer 1

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=$2\times x$=$2x$

Height of box= 4 inches greater than width= $x+4$

Volume of box= 48 cubic inches

We know that the formula of volume of cuboid= $length\times breadth\times height$

Apply the formula

Volume of box= $x\times 2x\times (x+4)$

Volume of cube = $2x^2(x+4)$

$2x^2(x+4)=48$

$x^2(x+4)=24$

$x^3+x^2-24$

Apply inspection method to solve the equation

Put $x=0$

Then we get $-24\neq0$

Hence, x=0 is not the solution of x

Put x=1 in the equation then we get

$-22\neq 0$

Hence x=1 is not the solution of equation.

Put x=2 then we get

$(2)^3+(4)^2-24$

8+16-24=0

Hence, x=2 is the solution of equation .

$(x-2)(x^2+6x+12)$=0

Now substitute equation $x^2+6x+12$=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 $x^2+6x+12=0$ are imaginary.

Lenght , widht and height are dimensions of box therefore, imaginary value are not possible.

Hence,$x=2$ 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 = $2\times2$=4inches

Height of box=$x+4$=2+4=6 inches