Question

Elena is making an open top box by cutting squares out of the corners of a piece of paper that is 11 inches wide and 17 in long and then folding up the sides if the side length of a square cut outs RX in in the volume of the box is given by $v(x) = x(11 - 2x)(17 - 2x)$what is a reasonable domain for V of x?approximately which value of x will give her a box with the greatest volume round to the nearest whole numberfor approximately which values of X is the volume of the Box increasing round to the nearest whole number

Answer 1

The expression for the volume of the box is:

$v(x)=x(11-2x)(17-2x)$

Mathematically, there is no restriction for the values of x, but phisically we know that x is a length and has a positive value, so x>0.

Also, we know that x can not be largest than half of the width, that is the smallest dimension of the piece of paper.

As the width is 11, we then know that x is smaller than 11/2=5.5.

In conclusion, the domain for x is:

$0To calculate the maximum volume for the box we have to derive the volume function and equal to zero:[tex]\begin{gathered} v(x)=x(11-2x)(17-2x) \\ v(x)=x(11\cdot17-11\cdot2x-2x\cdot17+4x^2) \\ v(x)=x(4x^2-56x+187) \\ v(x)=4x^3-56x^2+187x \end{gathered}$$\begin{gathered} \frac{dv}{dx}=4(3x^2)-56(2x)+187(1)=0 \\ 12x^2-112x+187=0 \\ x=\frac{-(-112)\pm\sqrt[]{(-112)^2-4\cdot12\cdot187}}{2\cdot12} \\ x=\frac{112\pm\sqrt[]{12544-8976}}{24} \\ x=\frac{112\pm\sqrt[]{3568}}{24} \\ x=\frac{112}{24}\pm\frac{59.73}{24} \\ x=4.67\pm2.49 \\ x_1=4.67-2.49=2.18\approx2 \\ x_2=4.67+2.49=7.16\approx7 \end{gathered}$

The solutions are x=2 and x=7 approximately.

Because of our domain definition, we know that x=7 is not a valid solution, so the value of x that maximizes the volume is x=2.

The volume for x=0 is 0. Then, it will increase its value until x=2, where it reaches the maximum volume. From x=2 to x=5.5, the volume decrease until reaching v=0 at x=6.5.

Answer:

Domain: 0

Value of x that maximizes the volume: x=2.

From x=0 to x=2 the volume of the box increases.